NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:41 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 3 Trigonometric Functions is considered a very important chapter for practical use, and application in various different fields and also for the exams. NCERT Exemplar Class 11 Maths chapter 3 solutions give a brief procedure and explanation about angles along with degree and radian measure. It also establishes a relationship between radian and real numbers, and radian and degree measure. Class 11 Maths NCERT Exemplar solutions chapter 3 also covers a variety of questions relating to a conversion of degrees to radian and vice versa through established relation between them through the use of notational convention.

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NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3
Important Notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions
Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3
What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?
NCERT Solutions for Class 11 Mathematics Chapters
Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3

Question:1

Prove that
$\\ \frac{tanA+secA-1}{tanA-secA+1} =\frac{1+sinA}{cosA}$

Answer:

$\\ L.H.S=\frac{tanA+secA-1}{tanA-secA+1} \\\\ =\frac{tanA+secA- \left( \sec ^{2}A-\tan ^{2}A \right) }{tanA-secA+1} \\\\ =\frac{tanA+secA- \left[ \left( secA+tanA \right) \left( secA-tanA \right) \right] }{tanA-secA+1}\\ \\ = \frac{ \left( secA+tanA \right) \left[ 1- \left( secA-tanA \right) \right] }{tanA-secA+1} \\\\$

$\\ = \frac{ \left( secA+tanA \right) \left[ 1-secA+tanA \right] }{tanA-secA+1} \\ \\ =secA+tanA \\ \\ =\frac{1}{cosA}+\frac{sinA}{cosA} \\ \\ =\frac{1+sinA}{cosA}=R.H.S \\ \\$

Question:2

If $[2sin \alpha / (1+cos \alpha +sin \alpha )] = y$ , then prove that $[(1- cos \alpha +sin \alpha ) / (1+sin \alpha )]$ is also equal to y.

$\left[\mathrm{Hint}: \text { Express } \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\right]$

Answer:

$\\y=\frac{2sin \alpha }{1+cos \alpha +sin \alpha }=\frac{2sin \alpha }{1+cos \alpha +sin \alpha }\ast\frac{1+sin \alpha -cos \alpha }{1+sin \alpha -cos \alpha } \\\\ =\frac{2sin \alpha \left( 1-cos \alpha +sin \alpha \right) }{ \left( 1+sin \alpha \right) ^{2}-\cos ^{2} \alpha } \\\\ =\frac{2sin \alpha \left( 1-cos \alpha +sin \alpha \right) }{1+\sin ^{2} \alpha +2sin \alpha -\cos ^{2} \alpha } \\\\$
$\\ =\frac{2sin \alpha \left( 1-cos \alpha +sin \alpha \right) }{ \left( 1-\cos ^{2} \alpha \right) +\sin ^{2} \alpha +2sin \alpha } \\\\ =\frac{2sin \alpha \left( 1-cos \alpha +sin \alpha \right) }{2sin^{2} \alpha +2sin \alpha } \\\\ = \frac{2sin \alpha \left( 1-cos \alpha +sin \alpha \right) }{2sin \alpha \left( 1+sin \alpha \right) }= \frac{ \left( 1-cos \alpha +sin \alpha \right) }{ \left( 1+sin \alpha \right) }=y \\\\$

Question:3

If $m sin \theta = n sin ( \theta + 2 \alpha )$ , then prove that
$tan ( \theta + \alpha ) cot \alpha = (m + n)/(m - n)$
[Hints: Express $sin( \theta + 2 \alpha ) / sin \theta = m/n$ and apply componendo and dividend]

Answer:

$\\Given \ that \ \ m sin \theta =nsin \left( \theta +2 \alpha \right) ~ \\\\ ~~\frac{\sin \left( \theta +2 \alpha \right) }{sin \theta }=\frac{m}{n}~~ \\\\ ~~\frac{\sin \left( \theta +2 \alpha \right) +sin \theta }{\sin \left( \theta +2 \alpha \right) -sin \theta }=\frac{m+n}{m-n}~~~ \\\\$

$\frac{2 \sin \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right)}=\frac{m+n}{m-n}$

$\tan \left( \theta + \alpha \right) .cot \alpha =\frac{m+n}{m-n} \\\\$

Question:4

If $\cos (\alpha+\beta)=\frac{4}{5} \text { and } \sin (\alpha-\beta)=\frac{5}{13}$ where α lie between 0 and π/4, find value of tan 2α
[Hint: Express $tan 2 \: \: \alpha \: \: as\: \: tan ( \alpha + \beta + \alpha - \beta ]$

Answer:

$\\ \cos \left( \alpha + \beta \right) =\frac{4}{5}\text{~ so,}\tan \left( \alpha + \beta \right) =\frac{3}{4}~~~~~ \\\\ \text{~~~~ And}\sin \left( \alpha - \beta \right) =\frac{5}{13}~~\tan \left( \alpha - \beta \right) =\frac{5}{12}~ \\\\ tan2 \alpha =\tan \left[ \alpha + \beta + \alpha - \beta \right] ~~ \\\\$
$\\ =\tan \left[ \left( \alpha + \beta \right) + \left( \alpha - \beta \right) \right] ~ \\\\ =\frac{\tan \left( \alpha + \beta \right) +\tan \left( \alpha - \beta \right) }{1-\tan \left( \alpha + \beta \right) \tan \left( \alpha - \beta \right) } \\\\$

$\\ =~ \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\ast\frac{5}{12}}=\frac{56}{33}~ \\\\$

Question:5

If $tan x = b/a$ then find the value of $\sqrt {\frac{a+b}{a-b}}+ \sqrt {\frac{a-b}{a+b}}$

Answer:

$\\tanx=\frac{b}{a}~~~~~ \\\\ \sqrt {\frac{a+b}{a-b}}+ \sqrt {\frac{a-b}{a+b}}= \frac{a+b+a-b}{\sqrt { \left( a-b \right) \left( a+b \right) }} \\\\ =\frac{2a}{\sqrt {a^{2}-b^{2}}} \\\\ =\frac{2a}{a\sqrt {1-\frac{b^{2}}{a^{2}}}} \\\\$

$\\ = \frac{2}{\sqrt {1-\tan ^{2}x}}=\frac{2}{\sqrt {1- \left( \sin ^{2}x \right) / \left( \cos ^{2}x \right) }}=2cosx /\left( \sqrt {cos2x} \right) \\\\$

Question:6

Prove that $\cos \theta \cos \frac{\theta}{2} - \cos 3 \theta \cos \frac{9\theta}{2} = \sin 7 \theta \sin 4 \theta$
$[\text{Hint:Express L.H.S. }=\frac{1}{2}[2cos \theta cos \theta /2 - 2cos 3 \theta cos 9 \theta / 2]$

Answer:

$\\L.H.S=cos \theta \cos \left( \frac{ \theta }{2} \right) - cos3 \theta \cos \left( \frac{9 \theta }{2} \right) ~~ \\\\ =\frac{1}{2} \left[ 2cos \theta \cos \left( \frac{ \theta }{2} \right) \right] - \frac{1}{2} \left[ 2cos3 \theta \cos \left( \frac{9 \theta }{2} \right) \right] \\\\ =\frac{1}{2} \left[ \cos \left( \theta +\frac{ \theta }{2} \right) +\cos \left( \theta - \frac{ \theta }{2} \right) \right] - \frac{1}{2} \left[ \cos \left( 3 \theta +\frac{9 \theta }{2} \right) +\cos \left( 3 \theta - \frac{9 \theta }{2} \right) \right] \\\\$

$=\frac{1}{2}\left[\cos \frac{3 \theta}{2}+\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}-\cos \left(-\frac{3 \theta}{2}\right)\right]$

$=\frac{1}{2} \left[ \cos \frac{3 \theta }{2}+\cos \frac{ \theta }{2} - \cos \frac{15 \theta }{2} - \cos \frac{3 \theta }{2} \right] ~ \\\\$

$\\ =\frac{1}{2} \left[ \cos \frac{ \theta }{2} - \cos \frac{15 \theta }{2} \right] ~ \\\\ =\frac{1}{2} \left[ - 2\sin \left( \frac{\frac{ \theta }{2}+\frac{15 \theta }{2}}{2} \right) \sin \left( \frac{\frac{ \theta }{2} - \frac{15 \theta }{2}}{2} \right) \right] \\\\$

$=\frac{1}{2} \left [ - 2\sin 4 \theta \sin \left( - \frac{7 \theta }{2} \right) \right ]$
$= - \sin \left( 4 \theta \right) \sin \left( - \frac{7 \theta }{2} \right) =\sin \left( 4 \theta \right) \sin \left( \frac{7 \theta }{2} \right) \\\\$

Question:7

If $a \cos \theta + b \sin \theta = \: \: m\: \: and \: \: a \sin \theta - b cos \theta = n$ , then show that $m^2+n^2=a^2+b^2$

Answer:

$\\a~cos \theta +b~sin \theta =m~~ a sin \theta - b cos \theta =n \\\\ ~ R.H.S=m^{2}+n^{2}= \left( a cos \theta +b sin \theta \right) ^{2}+ \left( a sin \theta - b cos \theta \right) ^{2}~ \\\\ =a^{2}\cos ^{2} \theta +b^{2}\sin ^{2} \theta +2ab sin \theta cos \theta +a^{2}\sin ^{2} \theta +b^{2}\cos ^{2} \theta - 2ab sin \theta cos \theta =a^{2} \left( \cos ^{2} \theta +\sin ^{2} \theta \right) +b^{2} \left( \cos ^{2} \theta +\sin ^{2} \theta \right) \\\\ =a^{2}+b^{2}~ \\\\$

Question:8

Find the value of $tan 22 ^{\circ} 30^{'}$ .

$\begin{aligned} &\text { [Hint: Let } \theta=45^{\circ} \text { , use }\\ &\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=\frac{\sin \theta}{1+\cos \theta} \end{aligned}]$

Answer:

$\\ Let~22^{0}30^{'}=\frac{ \theta }{2}~~~\tan 22^{0}30^{'}=\tan \frac{ \theta }{2}~ \\\\ =\frac{\sin \frac{ \theta }{2}}{\cos \frac{ \theta }{2}}= \frac{2\sin \frac{ \theta }{2}\cos \frac{ \theta }{2}}{2\cos ^{2}\frac{ \theta }{2}} \\\\ = \frac{sin \theta }{1+cos \theta }~~~~~ \\\\ Putting \theta =45 \\\\ \frac{sin \theta }{1+cos \theta }=\frac{\frac{1}{\sqrt {2}}}{1+\frac{1}{\sqrt {2}}} \\\\ =\frac{1}{\sqrt {2}+1}= \left( \sqrt {2~} - 1 \right) \left[ On rationalising \right] \\\\$

Question:9

Prove that $sin 4A = 4sinA cos\textsuperscript{3}A - 4 cosA sin\textsuperscript{3}A$ .

Answer:

$\\L.H.S\sin 4A=\sin \left( A+3A \right) ~ =sinA\cos 3A+\cos A\sin 3A~ \\\\ =\sin A \left( 4\cos ^{3}A - 3\cos A \right) +\cos A \left( 3 sinA - 4\sin ^{3}A \right) \\\\ =4\sin A\cos ^{3}A - 3 sinA\cos A+3 sinA\cos A - 4\cos A\sin ^{3}A \\\\ =4 sinA\cos ^{3}A - 4 cosA\sin ^{3}A=R.H.S \\\\$

Question:10

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$ , then prove that $m\textsuperscript{2} - n\textsuperscript{2} = 4 \sin\theta \tan\theta$
[Hint: $m + n = 2\tan \theta , m - n = 2 \sin \theta$ , then use $[m\textsuperscript{2} - n^2] = (m + n)(m - n)]$

Answer:

$\\\tan \theta +\sin \theta =m~and \tan \theta - \sin \theta =n \\\\ L.H.S=m^{2} - n^{2}= \left( m+n \right) \left( m - n \right) \\\\ = \left[ \left( \tan \theta +\sin \theta \right) + \left( \tan \theta - \sin \theta \right) \right] .~ \left[ \left( \tan \theta +\sin \theta \right) - \left( \tan \theta - \sin \theta \right) \right] \\\\ = \left( \\tan \theta +\sin \theta +\tan \theta - \sin \theta \right) . \left( \\tan \theta +\sin \theta - \tan \theta +\sin \theta \right) \\\\ =2 \tan \theta . 2\sin \theta =4 \sin \theta \tan \theta =R.H.S \\\\$

Question:11

If $tan (A + B) = p, tan (A - B) = q$ , then show that $tan 2A = (p + q) / (1 -pq)$ .
[Hint: Use $2A = (A + B) + (A - B)$ ]

Answer:

$\\ \tan \left( A+B \right) =p,tan \left( A - B \right) =q~~~ \\\\ ~\\tan 2A=tan \left( A+B+A - B \right) =tan \left[ \left( A+B \right) + \left( A - B \right) \right] \\\\ = \frac{\\tan \left( A+B \right) +\\tan \left( A - B \right) }{1 - \\tan \left( A+B \right) .\\tan \left( A - B \right) } \\\\ =\frac{p+q}{1 - pq} \\\\$

Question:12

If $\cos \alpha + \cos \beta = 0 = \sin \alpha + \sin \beta$ , then prove that $\cos 2 \alpha + \cos 2 \beta = -2\cos ( \alpha + \beta )$ . [Hint: $(\cos \alpha + \cos \beta )\textsuperscript{2} - (\sin \alpha + \sin \beta )\textsuperscript{2} = 0$ ]

Answer:

Given that: $\cos \alpha +\cos \beta =0~~ and \sin \alpha +\sin \beta =0 \\\\$

$\\ ~So, \left( \cos \alpha +\cos \beta \right) ^{2} - \left( \sin \alpha +\sin \beta \right) ^{2}=0 \\\\ ~~ \left( \\cos ^{2} \alpha +\\cos ^{2} \beta +2 \cos \alpha \cos \beta \right) - \left( \sin^{2} \alpha +\\sin ^{2} \beta +2\sin \alpha \sin \beta \right) =0 \\\\ ~ \left( \\cos ^{2} \alpha - \\sin ^{2} \alpha \right) + \left( \\cos ^{2} \beta - \\sin ^{2} \beta \right) +2 \left( \cos \alpha \cos \beta - \sin \alpha \sin \beta \right) =0 \\\\ \\cos 2 \alpha +cos 2 \beta +2 \cos \left( \alpha + \beta \right) =0 \\\\ \\cos 2 \alpha +\cos2 \beta = - 2 \cos \left( \alpha + \beta \right) \\\\$

Question:13

If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$ , then show that $\frac{\tan x}{\tan y}=\frac{a}{b}$ [Hint: Use componendo and Dividendo]

Answer:

$\\\frac{\sin \left( x+y \right) }{\\sin \left( x - y \right) }=\frac{a+b}{a - b}~~~~ \\\\ ~~\frac{\\sin \left( x+y \right) +\\sin \left( x - y \right) }{\\sin \left( x - y \right) - \\sin \left( x - y \right) }=\frac{a+b+a - b}{a+b - a+b} \\\\$

$\\ \frac{2 \sin \frac{x+y+x - y}{2} \cos \frac{x+y - x+y}{2}}{2 \cos \frac{x+y+x - y}{2} \sin \frac{x+y - x+y}{2}} =\frac{2a}{2b} \\\\$

$\\\frac{\sin x\cos y}{\cos x\sin y}=\frac{a}{b} \\\\$

$\\ \frac{\tan x}{\tan y}=\frac{a}{b} \\\\$

Question:14

If $\tan \theta =\frac{\sin \alpha - \cos \alpha }{\sin \alpha +\cos \alpha }$ then show that $\\\sin \alpha + \cos \alpha = \surd 2 \cos \theta$
[Hint: Express $\tan \theta = \tan( \alpha - \pi / 4) \\$ , $\theta = \alpha - \pi /4$ ]

Answer:

$\\ \tan \theta =\frac{\sin \alpha - \cos \alpha }{\sin \alpha +\cos \alpha }\\ \\\\ \tan \theta = \frac{\frac{\sin \alpha - \cos \alpha }{\cos \alpha }}{\frac{\sin \alpha +\cos \alpha }{\cos \alpha }}=\frac{\tan \alpha - 1}{\tan \alpha +1} \\\\ \\= \left( \tan \alpha - \\tan \frac{ \pi }{4} \right) / \left( 1+\\tan \frac{ \pi }{4}\tan \alpha \right) ~~~~ \\\\ \tan \theta =tan \left( \alpha - \frac{ \pi }{4} \right) ~~~ \\\\$

$\\ \theta = \alpha - \frac{ \pi }{4} \\\\$

$\cos \theta=\cos \left(\alpha-\frac{\pi}{4}\right)$

$\\ \cos \theta =\cos \alpha cos \frac{ \pi }{4}+\sin \alpha sin \frac{ \pi }{4}~ \\\\ ~ \cos \theta =\cos \alpha .\frac{1}{\sqrt {2}}+\sin \alpha .\frac{1}{\sqrt {2}}~~ \\\\$

$~\sqrt {2} \cos \theta =\cos \alpha +\sin \alpha \\\\$

Hence, proved

Question:15

If $\sin \theta + \cos \theta = 1$ , then find the general value of $\theta$

Answer:

Given that $\sin \theta +\cos \theta =1 \: \: \text{Dividing both sides by} \sqrt {1^{2}+1^{2}}= \sqrt {2} \\\\$

$\\ ~~\frac{1}{\sqrt {2}}\sin \theta +\frac{1}{\sqrt {2~}}\cos \theta =\frac{1}{\sqrt {2}}~ \\\\ ~\\cos \left( \theta - \frac{ \pi }{4} \right) =\cos \frac{ \pi }{4}~ \\\\ ~~~ \theta - \frac{ \pi }{4}=2n \pi \pm \frac{ \pi }{4}~,~n \epsilon Z \\\\$

$\\ \theta =2n \pi \pm \frac{ \pi }{4}+\frac{ \pi }{4} \\\\ ~~~ \theta =2n \pi +\frac{ \pi }{4}+\frac{ \pi }{4}~~~or \theta =2n \pi - \frac{ \pi }{4}+\frac{ \pi }{4}~~~ \\\\ ~ \theta =2n \pi +\frac{ \pi }{2}~ or \theta =2n \pi , n \epsilon Z \\\\$

Question:16

Find the most general value of $\theta$ satisfying the equation $\tan \theta = -1 and\: \: \cos \theta = 1/ \sqrt 2$

Answer:

Given that

$\\\tan \theta = - 1 and \cos \theta =\frac{1}{\sqrt {2}}~ \\\\ ~ \cos \theta \text{ is positive and } \tan \theta \: \text{is negative in fourth quadrant} \tan \theta = - 1 \\\\ ~~ \tan \theta =\tan \left( - \frac{ \pi }{4} \right) ~ \\\\ ~ \tan \theta =\tan \left( 2 \pi - \frac{ \pi }{4} \right) \\\\$

$\\ ~ \tan \theta =\tan \frac{7 \pi }{4}~ \\\\ \theta =\frac{7 \pi }{4}~~ \\\\ \cos \theta =\frac{1}{\sqrt {2}}~ \\\\ ~ \cos \theta =\cos \frac{ \pi }{4}~~~~~ \\\\$

$\\ ~~~ \cos \theta =\cos \left( 2 \pi - \frac{ \pi }{4} \right) ~ \\\\ \cos \theta =\cos \frac{7 \pi }{4}~~ \\\\ ~~~~ \theta =\frac{7 \pi }{4}~~~ \\\\ ~ \text{So, general solution is } \theta =2n \pi +\frac{7 \pi }{4} \\\\$

Question:17

If $\cot \theta + \tan \theta = 2 cosec \theta$ , then find the general value of $\theta$ .

Answer:

$\text{Given that}\cot \theta +tan \theta =2 cosec \theta ~~~ \\\\$

$\\ ~~\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=\frac{2}{\sin \theta }~~ \\\\ ~~~~\frac{\\sin ^{2} \theta +\\cos ^{2} \theta }{\sin \theta \cos \theta }=\frac{2}{\sin \theta }~~~~~~ \\\\ ~~~~~\frac{1}{\sin \theta \cos \theta }=\frac{2}{\sin \theta }~~~~~~~ \\\\ ~2\sin \theta \cos \theta =\sin \theta \\\\$
$\\ ~ \sin \theta \left( 2\cos \theta - 1 \right) =0 \\\\ ~\sin \theta =0~or\: 2\cos \theta - 1=0 \: or\: \cos \theta =\frac{1}{2}~~~~~~ \\\\ ~~Now,~\sin \theta =0\Rightarrow \theta =n \pi , n \epsilon Z \\\\ ~ \cos \theta =\frac{1}{2}\Rightarrow \cos \theta =\cos \frac{ \pi }{3}~~ \\\\ ~ \theta =2n \pi \pm \frac{ \pi }{3}~ \\\\$
Hence, general values of $\theta$ is $2n \pi \pm \frac{ \pi }{3}~ and \: \: n \pi ,n \epsilon Z \\\\$

Question:18

If $2\sin\textsuperscript{2} \theta = 3\cos \theta , where \: \: 0 \leq \theta \leq 2 \pi$ , then find the value of θ.

Answer:

$\\ \text{Given that }2\sin ^{2} \theta =3\cos \theta ~ \\\\ ~2 \left( 1 - \cos ^{2} \theta \right) =3~\cos \theta \\\\ 2 - 2\cos ^{2} \theta - 3~\cos \theta =0~ \\\\ 2\cos ^{2} \theta +3\cos \theta - 2=0~ \\\\$

$\left( \cos \theta +2 \right) \left( 2\cos \theta - 1 \right) =0~~~ \left[ \text{On factorising} \right] ~~~ \\\\$

$\cos \theta +2=0 or 2\cos \theta - 1=0 \\\\$

$\\cos \theta \neq - 2 ~ \\\\$

$So, 2\cos \theta - 1=0,\cos \theta =\frac{1}{2}~ \\\\$
$\\ or \theta =\frac{ \pi }{3}or~ 2 \pi - \frac{ \pi }{3} \\\\ \theta =\frac{ \pi }{3}or~~\frac{5 \pi }{3} \\\\$

Question:19

If $sec x \cos 5x + 1 = 0, where 0 < x \leq \pi /2$ , then find the value of x.

Answer:

$\\ \text{Given that } \ \sec x\cos 5x+1=0 \\ ~\frac{1}{\cos x~}.\cos 5x+1=0 \\ \\cos 5x+\cos x=0 \\\\ \text{~ 2}\cos \left( \frac{5x+x}{2} \right) \cos \left( \frac{5x - x}{2} \right) =0 \\\\$
$\\ ~\cos 3x.\cos 2x=0 ~ \\\\ ~\cos 3x=0~or\cos 2x=0 ~ \\\\ 3x=\frac{ \pi }{2}~ or 2x=\frac{ \pi }{2}~~~ \\\\ ~ x=\frac{ \pi }{6} or x=\frac{ \pi }{4}~~~ \\\\$

Question:20

If $\sin ( \theta + \alpha ) = a \ and \sin ( \theta + \beta ) = b$ , then prove that $\cos 2( \alpha - \beta ) - 4ab \cos ( \alpha - \beta ) = 1 - 2a\textsuperscript{2} - 2b\textsuperscript{2}$

Answer:

$\\ \text{Given that }\sin \left( \theta + \alpha \right) =a~ and\sin \left( \theta + \beta \right) =b~ \ldots \ldots \ldots \ldots \left( i \right) \\\\ \cos \left( \alpha - \beta \right) =\cos \left[ \theta + \alpha - \theta - \beta \right] \\\\ =\cos \left[ \left( \theta + \alpha \right) - \left( \theta + \beta \right) \right] \\\\$

$\\ =\cos \left( \theta + \alpha \right) \cos \left( \theta + \beta \right) +\sin \left( \theta + \alpha \right) \sin \left( \theta + \beta \right) \\\\ = \sqrt {1 - \sin ^{2} \left( \theta + \alpha \right) }.\sqrt {1 - \sin ^{2} \left( \theta + \beta \right) }+\sin \left( \theta + \alpha \right) \sin \left( \theta + \beta \right) ~ \\\\ =\sqrt { \left( 1 - a^{2} \right) \left( 1 - b^{2} \right) }+ab \\\\ ~~\cos \left( \alpha - \beta \right) =ab+ \sqrt {1 - a^{2} - b^{2}+a^{2}b^{2}}~ \\\\$

$\\ ~Now,\cos 2 \left( \alpha - \beta \right) - 4ab\cos \left( \alpha - \beta \right) ~ \) = 2\cos ^{2} \left( \alpha - \beta \right) - 1 - 4ab\cos \left( \alpha - \beta \right) \\\\ =2 \left[ ab+\sqrt {1 - a^{2} - b^{2}+a^{2}b^{2}}\right]^2 - 1 - 4ab \left[ ab+\sqrt {1 - a^{2} - b^{2}+a^{2}b^{2}}\right]~ \\\\$

$\\\\ =2 \left[ a^{2}b^{2}+1 - a^{2} - b^{2}+a^{2}b^{2}+2ab\sqrt {1 - a^{2} - b^{2}+a^{2}b^{2}} \right] - 1 - 4a^{2}b^{2} - 4ab\sqrt {1 - a^{2} - b^{2}+a^{2}b^{2}}~~ \\\\ =1 - 2a^{2} - 2b^{2} \\\\$

Question:21

If $\cos ( \theta + \phi ) = m \cos ( \theta - \phi )$ , then prove that $\tan \theta = ((1 - m)/(1 + m)) cot \phi$
[Hint: Express $\cos ( \theta + \phi )/ \cos ( \theta - \phi ) = m/1$ and apply Componendo and Dividendo]

Answer:

$\\ \text{Given that}\cos \left( \theta + \phi \right) =m\cos \left( \theta - \phi \right) ~~ \\\\ ~~~\frac{\cos \left( \phi + \theta \right) }{\cos \left( \theta - \phi \right) }=\frac{m}{1} \\\\$

$\text{~~Using~componendo~and dividend rule, }\frac{\cos \left( \phi + \theta \right) +\cos \left( \theta - \phi \right) }{\cos \left( \phi + \theta \right) - \cos \left( \theta - \phi \right) } =\frac{m+1}{m - 1}~ \\\\$
$\\ ~~ \frac{\cos \theta \cos \phi }{ - \sin \theta \sin \phi }=\frac{m+1}{m - 1}~~ \\\\ - cot \theta cot \phi =\frac{m+1}{m - 1}~ \\\\ ~\frac{ - cot \phi }{\tan \theta }=\frac{m+1}{m - 1} \\\\ ~~ \tan \theta \left( 1+m \right) = \left( 1 - m \right) cot \phi ~ \\\\$

$\\ ~ \tan \theta =\frac{1 - m}{1+m}cot \phi \\\\$

Question:22

Find the value of the expression
$=3 \left[ \sin ^{4} \left( \frac{3 \pi }{2} - \alpha \right) +\sin ^{4} \left( 3 \pi + \alpha \right) \right] - 2 \left[ \sin ^{6} \left( \frac{ \pi }{2}+ \alpha \right) +\sin ^{6} \left( 5 \pi - \alpha \right) \right]$

Answer:

$\\ =3 \left[ \sin ^{4} \left( \frac{3 \pi }{2} - \alpha \right) +\sin ^{4} \left( 3 \pi + \alpha \right) \right] - 2 \left[ \sin ^{6} \left( \frac{ \pi }{2}+ \alpha \right) +\sin ^{6} \left( 5 \pi - \alpha \right) \right] \\\\ =3 \left[ \cos ^{4} \alpha +\sin ^{4} \left( \pi + \alpha \right) \right] - 2 \left[ \cos ^{6} \alpha +\sin ^{6} \left( \pi - \alpha \right) \right] ~ \\\\ =3 \left[ \cos ^{4} \alpha +\sin ^{4} \alpha \right] - 2 \left[ \cos ^{6} \alpha +\sin ^{6} \alpha \right] \\\\ =3 \left[ \cos ^{4} \alpha +\sin ^{4} \alpha +2\sin ^{2} \alpha \cos ^{2} \alpha - 2\sin ^{2} \alpha \cos ^{2} \alpha \right] - 2 \left[ \left( \cos ^{2} \alpha +\sin ^{2} \alpha \right) ^{3} - 3\cos ^{2} \alpha \sin ^{2} \alpha \left( \cos ^{2} \alpha +\sin ^{2} \alpha \right) \right] \\\\$

$\\=3 \left[ \left( \cos ^{2} \alpha +\sin ^{2} \alpha \right) ^{2} - 2\sin ^{2} \alpha \cos ^{2} \alpha \right] - 2 \left[ 1 - 3\cos ^{2} \alpha \sin ^{2} \alpha \right] \\\\ =3 \left[ 1 - 2\sin ^{2} \alpha \cos ^{2} \alpha \right] - 2 \left[ 1 - 3\cos ^{2} \alpha \sin ^{2} \alpha \right] \\\\ =3 - 6\sin ^{2} \alpha \cos ^{2} \alpha - 2+6\cos ^{2} \alpha \sin ^{2} \alpha \\\\ =3 - 2=1 \\\\$

Question:23

If $a \cos2 \theta + b \sin 2 \theta = c$ has α and β as its roots, then prove that $\tan \alpha + \tan \beta = 2b/(a + c)$ [Hint: Use the identities $\cos 2 \theta = (( 1 - \tan\textsuperscript{2} \theta )/(1 + \tan^2 \theta ) and \sin 2 \theta = 2\tan \theta /(1 + \tan^2 \theta )$ ]

Answer:

$\\ \text{Given that a}\cos 2 \theta +b\sin 2 \theta =c \ldots \ldots \ldots \left( i \right) ~~~ \\\\ ~ a - a\tan^{2} \theta +2b \tan \theta =c \left( 1+\tan ^{2} \theta \right) ~ \\\\ ~ a - a\tan ^{2} \theta +2b \tan \theta =c+c \tan^{2} \theta \\\\ \left( a+c \right) \tan ^{2} \theta - 2b \tan \theta + \left( c - a \right) =0 \ldots .. \left( ii \right) ~ \\\\$
Since $\alpha \: \: and \: \: \beta$ are the roots of the equation i)we have tanα and tanβ are the roots of ii)
$\\ \tan \alpha +\tan \beta = \frac{ - \left( - 2b \right) }{a+c}~~~~ \left[ \text{Sum of roots of a quadratic equation} \right] ~~ \\\\ \tan \alpha +\tan \beta =2b/ \left( a+c \right) \\\\$

Question:24

If $x = sec \phi - \tan \phi \: \: and \: \: y = cosec \phi + cot \phi$ , then show that xy + x – y + 1 = 0. [Hint: Find $xy + 1$ and then show tan $x - y = -(xy + 1)$ ]

Answer:

$\\ x=sec \phi - \tan \phi ~~~y=cosec \phi +cot \phi \\\\ L.H.S=xy+x - y+1 \\\\ = \left( sec \phi - \tan \phi \right) \left( cosec \phi +cot \phi \right) + \left( sec \phi - \tan \phi \right) - \left( cosec \phi +cot \phi \right) +1 \\\\ = \left( \frac{1}{cosec \phi } - \frac{\sin \phi }{\cos \phi } \right) \left( \frac{1}{\sin \phi }+\frac{\cos \phi }{\sin \phi } \right) + \left( \frac{1}{\cos \phi } - \frac{\sin \phi }{\cos \phi } \right) - \left( \frac{1}{\sin \phi }+\frac{\cos \phi }{\sin \phi } \right) \\+1 \\\\$

$\\ = \left( \frac{1 - \sin \phi }{\cos \phi } \right) \left( \frac{1+\cos \phi }{\sin \phi } \right) +\frac{1 - \sin \phi }{\cos \phi } - \frac{1+\cos \phi }{\sin \phi }+1~~$

$=\frac{1 - \sin \phi +\cos \phi - \sin \phi \cos \phi }{\cos \phi \sin \phi }+\frac{\sin \phi - \sin ^{2} \phi - \cos \phi - \cos ^{2} \phi }{\cos \phi \sin \phi} +1 \\\\$

$\\ =\frac{1 - \sin \phi +\cos \phi - \sin \phi \cos \phi +\sin \phi - \sin ^{2} \phi - \cos \phi - \cos ^{2} \phi +\cos \phi \sin \phi }{\cos \phi \sin \phi } \\\\ =\frac{1 - 1}{\cos \phi \sin \phi }=0 \\\\ L.H.S= R.H.S \\\\$

Question:25

If θ lies in the first quadrant and cos θ = 8/17, then find the value of $\cos(30 ^{\circ} + \theta ) + \cos (45 ^{\circ} - \theta ) + \cos (120 ^{\circ} - \theta )$

Answer:

$\\ \text{Given that } \cos \theta =\frac{8}{17}~~~ \sin \theta =\sqrt {1 - \left( \frac{8}{17} \right) ^{2}}=\sqrt {1 - \frac{64}{289}}= \sqrt {\frac{289 - 64}{289}}\\= \pm \frac{15}{17}~ \\\\ But \: \: \theta \text{ lies in I quadrant and so} \sin \theta \text{ is positive.} \sin \theta =\frac{15}{17}~ \\\\ ~Now\cos \left( 30^{0}+ \theta \right) +\cos \left( 45^{0} - \theta \right) +\cos \left( 120^{0} - \theta \right) \\\\ =\cos30^{0}\cos \theta - \sin30^{0}\sin \theta +\cos45^{0}\cos \theta +\sin45^{0}\sin \theta +\cos120^{0}\cos \theta +\sin120^{0}\sin \theta \\\\$

$\\ =\frac{\sqrt {3}}{2}\cos \theta - \frac{1}{2}\sin \theta +\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta - \frac{1}{2}\cos \theta +\frac{\sqrt {3}}{2}\sin \theta \\\\ =\frac{\sqrt {3}}{2} \left( \cos \theta +\sin \theta \right) - \frac{1}{2} \left( \sin \theta +\cos \theta \right) +\frac{1}{\sqrt {2}} \left( \cos \theta +\sin \theta \right) \\\\ = \left( \frac{\sqrt {3}}{2} - \frac{1}{2}+\frac{1}{\sqrt {2}} \right) \left( \cos \theta +\sin \theta \right) \\\\ = \left( \frac{\sqrt {3} - 1}{2}+\frac{1}{\sqrt {2}} \right) \left( \frac{8}{17}+\frac{15}{17} \right) \\\\$

$\\ =\frac{\sqrt {3} - 1+\sqrt {2}~}{2}\ast\frac{23}{17}=\frac{23}{34} \left( \sqrt {3} - 1+\sqrt {2} \right) \\\\$

Question:26

Find the value of the expression $\cos\textsuperscript{4}( \pi /8) + \cos\textsuperscript{4}(3 \pi /8) + \cos\textsuperscript{4}(5 \pi /8) + \cos\textsuperscript{4}(7 \pi /8).$
[Hint: Simplify the expression to
$2\left(\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}\right)=2\left[\left(\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}\right)^{2}-2 \cos ^{2} \frac{\pi}{8} \cos ^{2} \frac{3 \pi}{8}\right]$

Answer:

$\\\cos^{4}\frac{ \pi }{8}+\cos ^{4}\frac{3 \pi }{8}+\cos ^{4}\frac{5 \pi }{8}+\cos ^{4}\frac{7 \pi }{8}~ \\\\ =\cos ^{4}\frac{ \pi }{8}+\frac{\cos ^{4}3 \pi }{8}+\cos ^{4} \left( \pi - \frac{3 \pi }{8} \right) +\cos ^{4} \left( \pi - \frac{ \pi }{8} \right) \\\\$

$\\ =\cos ^{4}\frac{ \pi }{8}+\cos ^{4} \frac{3 \pi }{8} +\cos ^{4}\frac{3 \pi }{8}+\cos ^{4}\frac{ \pi }{8} \\\\$
$\\ =2\cos ^{4}\frac{ \pi }{8}+2\cos ^{4}\frac{3 \pi }{8} \\\\ =2 \left[ \cos ^{4}\frac{ \pi }{8}+\cos ^{4}\frac{3 \pi }{8} \right] \\\\ =2 \left[ \cos ^{4}\frac{ \pi }{8}+\cos ^{4} \left( \frac{ \pi }{2} - \frac{ \pi }{4} \right) \right] =2 \left[ \cos ^{4}\frac{ \pi }{8}+\sin ^{4}\frac{ \pi }{8} \right] \\\\$

$\\ = 2 \left[ \left( \cos ^{2}\frac{ \pi }{8}+\sin ^{2}\frac{ \pi }{8} \right) ^{2} - 2\sin ^{2}\frac{ \pi }{8}\cos ^{2}\frac{ \pi }{8} \right] \\\\ =2 \left[ 1 - 2\sin ^{2}\frac{ \pi }{8}\cos ^{2}\frac{ \pi }{8} \right] \\\\ =2 - \left( \sin \frac{ \pi }{4} \right) ^{2}~~ \\\\ =2 - \left( \frac{1}{\sqrt {2}} \right) ^{2} \\\\ =2 - \frac{1}{2}=\frac{3}{2} \\\\$

Question:27

Find the general solution of the equation
$5\cos\textsuperscript{2} \theta + 7\sin^2 \theta - 6 = 0$

Answer:

$\\5\cos ^{2} \theta +7\sin ^{2} \theta - 6=0~ \\\\ 5\cos ^{2} \theta +7 \left( 1 - \cos^{2} \theta \right) - 6=0 \\\\ 5\cos ^{2} \theta +7 - 7\cos ^{2} \theta - 6=0~ \\\\~~~ - 2\cos ^{2} \theta +1=0 \\\\$

$\\ ~2\cos ^{2} \theta =1 \\\\ \cos ^{2} \theta =\frac{1}{2}~ \\\\ ~\cos ^{2} \theta =\cos ^{2}\frac{ \pi }{4}~ \\\\ ~~\frac{1+\cos2 \theta }{2}=\frac{1+\frac{\cos \pi }{2}}{2}~ \\\\$

$\\ \cos 2 \theta =\cos \frac{ \pi }{2}~~~ \\\\ ~ 2 \theta =2n \pi \pm \frac{ \pi }{2} \\\\ ~~ \theta =n \pi \pm \frac{ \pi }{4}~ \\\\$

Question:28

Find the general solution of the equation $\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x$

Answer:

$\\ \sin x - 3 \sin 2x+\sin3x=\cos x - 3\cos 2x+\cos 3x~~ \\\\$

$\\ \left( \sin 3x+\sin x \right) - 3\sin 2x= \left( \cos 3x+\cos x \right) - 3\cos 2x \\\\ 2\sin \left( \frac{3x+x}{2} \right) .\cos \left( \frac{3x - x}{2} \right) - 3\sin 2x =2\cos \left( \frac{3x+x}{2} \right) .\cos \left( \frac{3x - x}{2} \right) - 3\cos 2x \\\\ \text{2}\sin 2x\cos x - 3\sin 2x=2\cos 2x.\cos x - 3\cos 2x~~ \\\\$

$\\ 2\sin 2x\cos x - 2\cos 2x\cos x=3\sin 2x - 3\cos 2x~ \\\\ ~2\cos x \left( \sin 2x - \cos 2x \right) =3 \left( \sin2x - \cos 2x \right) \\\\ ~~ \left( \sin 2x - \cos 2x \right) \left( 2\cos x - 3 \right) =0 ~ \\\\ \sin 2x - \cos 2x=0 \\\\$
$\\ \text{ 2}\cos x - 3 \neq 0 \\\\ \frac{\sin 2x}{\cos2x} - 1=0 \\\\ \tan 2x=1\Rightarrow \tan 2x=\tan \left( \pi /4 \right) \\\\ ~ 2x= n \pi +\frac{ \pi }{4}~~ \\\\ ~~x=n \pi /2+ \pi /8 \\\\$

Question:29

Find the general solution of the equation $(\sqrt 3 - 1) \cos \theta + ( \sqrt 3 + 1) \sin \theta = 2$
[Hint: Put $Put \: \: \sqrt 3 - 1 = r \sin \alpha , \sqrt 3 + 1 = r \cos \alpha$ which gives $\tan \alpha = \tan(( \pi /4) - ( \pi /6)) \alpha = \pi /12$ ].

Answer:

$\\\left( \sqrt {3} - 1 \right) \cos \theta + \left( \sqrt {3}+1 \right) \sin \theta =2 \\\\ \text{~ put }\sqrt {3} - 1=r\sin \alpha ,~ \sqrt {3}+1=r\cos \alpha \\\\ \text{~squaring~and adding we get }r^{2}=3+1 - 2\sqrt {3}+3+1+2\sqrt {3}~~ \\\\ ~~r^{2}=8,~~ r=2\sqrt {2}~~ \\\\$

$\\ ~~ r\sin \alpha \cos \theta +r\cos \alpha \sin \theta =2 \\\\ \text{~ r} \left( \sin \alpha \cos \theta +\cos \alpha \sin \theta \right) =2 \\\\ ~2\sqrt {2}\sin \left( \alpha + \theta \right) =2~ \\\\ \sin \left( \alpha + \theta \right) =\frac{2}{2\sqrt {2}} \\\\ ~\sin \left( \theta + \alpha \right) =\sin \frac{ \pi }{4}~~ \\\\$

$\\ \alpha + \theta =n \pi + \left( - 1 \right) ^{n}\frac{ \pi }{4} \ldots \ldots . \left( i \right) ~ \\\\ \text{~ Now, }\frac{r\sin \theta }{r\cos \alpha }=\frac{\sqrt {3} - 1}{\sqrt {3}+1}~~ \\\\ \tan \alpha =\frac{\tan \frac{ \pi }{3} - \tan \frac{ \pi }{4}}{1+\tan \frac{ \pi }{4}\tan \frac{ \pi }{3}}~ \\\\ ~\tan \alpha =\tan \left( \frac{ \pi }{3} - \frac{ \pi }{4} \right) \\\\ ~~\tan \alpha =\tan \frac{ \pi }{12} \\\\$

$\\ \alpha =\frac{ \pi }{12}~ \\\\ \text{ Putting the value of } \alpha { in equation} \left( i \right) \text{~we get }\frac{ \pi }{12}+ \theta =n \pi + \left( - 1 \right) ^{n}.\frac{ \pi }{4}~~~~ \\\\ ~~~~~ \theta =n \pi + \left( - 1 \right) ^{n}.\frac{ \pi }{4} - \frac{ \pi }{12}~ \\\\$

Question:30

If $\sin \theta + cosec \theta = 2$ , then $\sin\textsuperscript{2} \theta + cosec\textsuperscript{2} \theta$ is equal to
A. 1
B. 4
C. 2
D. None of these

Answer:

The answer is the option (c).

$\\\sin \theta +cosec \theta =2 \\\\ ~ \left( \sin \theta +cosec \theta \right) ^{2}=2^{2}~~ \\\\ ~\sin ^{2} \theta +cosec^{2} \theta +2\sin \theta cosec \theta =4~ \\\\ ~\sin ^{2} \theta +cosec^{2} \theta +2\sin \theta cosec \theta =4~~ \\\\$

$\\ ~~~\sin ^{2} \theta +cosec^{2}~ \theta +2=4~~ \\\\ ~\sin ^{2} \theta +cosec^{2} \theta =2 \\\\$

Question:31

If $f(x) = \cos\textsuperscript{2}x + sec\textsuperscript{2}x$ , then

A. $f(x) < 1\\\\$

B. $f(x) = 1\\\\$

C. $2 < f(x) < 1\\\\$

D. $f(x) \geq 2\\\\$

[Hint: $A.M \geq G.M.$ ]

Answer:

The answer is the option (d)
$\\ f \left( x \right) =\cos ^{2}x+\sec ^{2}x~~ \\\\ We~know~that AM \geq GM \\\\ \frac{ \left( \cos^{2}x+\sec ^{2}x \right) }{2} \geq \sqrt {\cos ^{2}xsec^{2}~x}~~~~ \\\\ ~~\frac{ \left( \cos^{2}x+\sec ^{2}x \right) }{2} \geq 1~ \\\\ ~~\cos ^{2}x+\sec ^{2}x \geq 2~ \\\\ \text{~ f} \left( x \right) \geq 2 \\\\$

Question:32

If $\tan \theta = 1/2$ and $\tan \phi = 1/3,$ then the value of $\theta + \phi$ is

A. $\pi /6\\\\$

B. $\pi \\\\$

C. $0\\\\$

D. ${\pi /4} \\\\$

Answer:

The answer is the option (d)
$\\ \tan \left( \theta + \phi \right) =\frac{\tan \theta +\tan \phi }{1 - \tan \theta \tan \phi } \\\\ =\frac{\frac{1}{2}+\frac{1}{3}}{1 - \frac{1}{2}\ast\frac{1}{3}}~ \\\\ =\frac{\frac{5}{6}}{\frac{5}{6}}=1 \\\\ \tan \left( \phi + \theta \right) =\tan \frac{ \pi }{4}~ \\\\ ~~ \left( \theta + \phi \right) =\frac{ \pi }{4}~ \\\\$

Question:33

Which of the following is not correct?
A. $\sin \theta = - 1/5 \\\\$
B. $\cos \theta = 1\\\\$
C. $sec \theta = 1/2$
D. $\tan \theta = 20$

Answer:

The answer is the option (c)
$\\ \sin \theta = - \frac{1}{5}$ is correct since $- 1 \leq \sin \theta \leq 1 \\\\$
$\\ \cos \theta =1 \text{is true for} \theta =1 \\\\ ~ sec \theta = - \frac{1}{2}~~~~ \\\\ \cos \theta =2 \text{is not correct as }- 1 \leq \cos \theta \leq 1 \\\\$

Question:34

The value of $\tan 1 ^{\circ} \tan 2 ^{\circ} \tan 3 ^{\circ} \ldots \tan 89 ^{\circ} \: \: \: is\\\\$
A. 0
B. 1
C. $1/2$
D. Not defined

Answer:

The answer is the option (b).
$\\\text{Given that } \tan1^{0}~\tan2^{0} \ldots \ldots \ldots \ldots \ldots .\tan89^{0}~~ \\\\ =\tan1^{0}~\tan2^{0} \ldots \ldots \tan45^{0}\tan \left( 90 - 44^{0} \right) \tan \left( 90 - 43^{0} \right) \ldots .\tan \left( 90 - 1^{0} \right) \\\\ =\tan1^{0}\cot 1^{0}\tan 2^{0}\cot 2^{0} \ldots \ldots \ldots \ldots .\tan89^{0}\cot 89^{0} \\\\ =1.1 \ldots 1 \ldots \ldots 1.1=1 \\\\$

Question:35

The value of $\left (1 - \tan\textsuperscript{2}15\textsuperscript{o})/(1 + \tan\textsuperscript{2}15\textsuperscript{o} \right )$ is
A. 1

B. $\sqrt {3} \\\\$

C. $\sqrt 3/2\\\\$
D. 2

Answer:

The answer is the option (c).
$\\Given~that~~\frac{1 - \tan ^{2}15^{0}}{1+\tan ^{2}15^{0}}~~~~ \\\\ Let \: \: \theta =15^{0}~ 2 \theta =30^{0}~~ \\\\ ~ \cos2 \theta =\frac{1 - \tan ^{2} \theta }{1+\tan ^{2} \theta }~~ \\\\ ~\cos 30^{0}=\frac{1 - \tan ^{2}15^{0}}{1+\tan ^{2}15^{0}}=\frac{\sqrt {3}}{2} \\\\$

Question:36

The value of $\cos 1 ^{\circ} \cos 2 ^{\circ} \cos 3 ^{\circ} \ldots \cos 179 ^{\circ} \: \: is\\\\$
A. $1/ \sqrt 2\\\\$
B. 0
C. 1
D. -1

Answer:

The answer is the option (b).
$\\ \cos 1^{0}~\cos2^{0} \ldots \ldots \ldots \ldots \ldots .\cos179^{0} \\\\ = \cos1^{0}~\cos2^{0} \ldots \ldots \ldots \ldots \ldots .\cos 90^{0} \ldots \ldots .\cos179^{0} \] =0 \left( as~~\cos 90^{0}=0 \right) \\\\$

Question:37

If $\tan \theta = 3$ and $\theta$ lies in third quadrant, then the value of $\sin \theta \: \: is\\\\$

$\\A. \ 1/ \sqrt{ 10}\\\\ B. - 1/ \sqrt {10}\\\\ C. - 3/ \sqrt {10}\\\\ D. 3/ \sqrt {10} \\\\$

Answer:

The answer is the option (c).
$\\\tan \theta =3,~ \theta \text { lies in third quadrant, it is positive } \\\\ ~ \tan \theta =\frac{P}{B}=\frac{3}{1}~ \\\\ ~ Then, hypotenuse= \sqrt {3^{2}+1^{2}}=\sqrt {9+1}=\sqrt {10}~ \\\\ ~~ \sin \theta =\frac{3}{\sqrt {10}}~ where \theta \text{ lies in third quadrant} \\\\$

Question:38

The value of $\tan 75 ^{\circ} - cot 75 ^{\circ}$ is equal to
A. $2 \sqrt 3$
${B. 2 + \sqrt 3} \\\\$
${C. 2 - \sqrt 3 \\\\$
D. 1

Answer:

The answer is the option (a).
$\\ \tan 75 - \cot 75=\tan 75 - \cot \left( 90 - 15 \right) =\tan 75 - \tan 15 \\\\ =\frac{\sin 75}{\cos 75} - \frac{\sin 15}{\cos 15} \\\\ =\frac{ \left( \sin 75\cos 15 - \sin 15\cos 75 \right) }{\cos 75\cos 15}=\frac{\sin \left( 75 - 15 \right) }{\frac{1}{2} \times 2\cos 75\cos 15} \\\\ =\frac{2\sin 60}{\cos \left( 75+15 \right) +\cos \left( 75 - 15 \right) }=\frac{2\sin 60}{\cos 90+\cos 60} \\\\ =\frac{2 \times \frac{\sqrt {3}}{2}}{0+\frac{1}{2}}=2\sqrt {3} \\\\$

Question:39

Which of the following is correct?

A. $\sin 1 ^{\circ} > \sin 1\\\\$

B. $\sin 1 ^{\circ} < \sin 1\\\\$

C. $\sin 1 ^{\circ} = \sin 1\\\\$

D. $\sin 1^{\circ}=\frac{\pi}{18^{\circ}} \sin 1$

[Hint: $1 radian = 180 ^{\circ} \pi = 57 ^{\circ} 30'$ approx.]

Answer:

The answer is the option (b).
If $~ \theta$ increases then the value of $\sin \theta$ also increases. \\\\

So, $\sin1^{\circ}<\sin 1$
Hence, b is correct.

Question:40

If $\tan \alpha=\frac{\mathrm{m}}{\mathrm{m}+1}, \tan \beta=\frac{1}{2 \mathrm{m}+1}$ then $\alpha + \beta$ is equal to
$\\A. \frac{\pi}{2}\\\\ B.\frac{\pi}{3}\\\\ C.\frac{\pi}{c} \\\\ D.\frac{\pi}{4}$

Answer:

The answer is the option (d).
$\\ \tan \alpha =\frac{m}{m+1} \\\\ \tan \beta =\frac{m}{2m+1} \\\\ \tan \left( \alpha + \beta \right) =\frac{\tan \alpha +\tan \beta }{1 - \tan \alpha \tan \beta }\\\\=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1 - \frac{m}{m+1} \times \frac{1}{2m+1}}\\\\=\frac{\frac{2m^{2}+m+m+1}{ \left( m+1 \right) \left( 2m+1 \right) }}{\frac{ \left( m+1 \right) \left( 2m+1 \right) - m}{ \left( m+1 \right) \left( 2m+1 \right) }}\\\\\\=\frac{2m^{2}+2m+1}{2m^{2}+2m+m+1 - m}\\\\=\frac{2m^{2}+2m+1}{2m^{2}+2m+1}\\\\=1 \\\\ \tan \left( \alpha + \beta \right) =\tan \frac{ \pi }{4} \\\\ \alpha + \beta =\frac{ \pi }{4} \\\\$

Question:41

The minimum value of $3 \cos x + 4 \sin x + 8$ is
A. 5
B. 9
C. 7
D. 3

Answer:

The answer is the option (d).

$\\Let\ y=3\cos x+4\sin x+8 \\\\ y - 8= 3\cos x+4\sin x \\\\ \text{Minimum value of } y - 8= - \sqrt { \left( 3 \right) ^{2}+ \left( 4 \right) ^{2}}= - 5 \\\\ y=8 - 5=3 \\\\$

Hence, (d) is the correct option.

Question:42

The value of $\tan 3A - \tan 2A -\tan A$ is equal to
A. $\tan 3A \tan 2A \tan A$
B. $- \tan 3A \tan 2A \tan A$
C. $\tan A \tan 2A - \tan 2A \tan 3A - \tan 3A \tan A$
D. None of these

Answer:

The answer is the option (a).

$\\ \tan 3A=\tan \left( 2A+A \right) =\frac{\tan 2A+\tan A}{1 - \tan 2A\tan A} \\\\ \tan 3A \left( 1 - \tan 2A\tan A \right) =\tan 2A+\tan A \\\\ \tan 3A - \tan 3A\tan 2A\tan A=\tan 2A+\tan A \\\\ \tan 3A\tan 2A\tan A=\tan 3A - \tan 2A - \tan A \\\\$
Hence, a is correct.

Question:43

The value of $\sin (45 ^{\circ} + \theta ) - \cos (45 ^{\circ} - \theta )$ is
A. $2 \cos \theta$
B. $2\sin \theta$
C. 1
D. 0

Answer:

The answer is the option (d).
$\\ \sin \left( 45+ \theta \right) - \cos \left( 45 - \theta \right) \\\\ \sin \left( 45+ \theta \right) =\sin 45\cos \theta +\cos 45\sin \theta =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta \\\\ \cos \left( 45 - \theta \right) =\cos 45\cos \theta +\sin 45\sin \theta =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta \\\\ \sin \left( 45+ \theta \right) - \cos \left( 45 - \theta \right) =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta - \frac{1}{\sqrt {2}}\cos \theta - \frac{1}{\sqrt {2}}\sin \theta =0 \\\\$

Question:44

The value of $\cot \left( \frac{ \pi }{4}+ \theta \right) \cot \left( \frac{ \pi }{4} - \theta \right)$ is
A. –1
B. 0
C. 1
D. Not defined

Answer:

The answer is the option (c).

$\\ \cot \left( \frac{ \pi }{4}+ \theta \right) \cot \left( \frac{ \pi }{4} - \theta \right) =\frac{\cot \frac{ \pi }{4}\cot \theta - 1}{\cot \theta +\cot \frac{ \pi }{4}} \times \frac{\cot \frac{ \pi }{4}\cot \theta +1}{\cot \theta - \cot \frac{ \pi }{4}} \\\\ =\frac{\cot \theta - 1}{\cot \theta +1} \times \frac{\cot \theta +1}{\cot \theta - 1}=1 \\\\$
(c) is correct.

Question:45

$\cos 2 \theta \cos 2 \phi + \sin^2( \theta - \phi ) - \sin^2( \theta + \phi )$ is equal to
A. $\sin 2( \theta + \phi )$
B. $\cos 2( \theta + \phi )$
C. $\sin 2( \theta - \phi )$
D. $\cos 2( \theta - \phi )$
[Hint: Use $\sin2A - \sin2B = \sin (A + B) \sin (A - B)$ ]

Answer:

The answer is the option (b).
$\\ \cos 2 \theta \cos 2 \varnothing +\sin ^{2} \left( \theta - \varnothing \right) +\sin ^{2} \left( \theta + \varnothing \right) \\\\ since,~\sin ^{2}A - \sin ^{2}B=\sin \left( A+B \right) \sin \left( A - B \right) \\\\ =\cos 2 \theta \cos 2 \varnothing +\sin \left( \theta - \varnothing + \theta + \varnothing \right) \sin \left( \theta - \varnothing - \theta - \varnothing \right) \\\\ =\cos 2 \theta \cos 2 \varnothing - \sin 2 \theta \sin 2 \varnothing \\\\ since,~\cos x\cos y - \sin x\sin y=\cos \left( x+y \right) \\\\ =\cos \left( 2 \theta +2 \varnothing \right) \\\\ =\cos 2 \left( \theta + \varnothing \right) \\\\$
Hence, the correct option is (b).

Question:46

The value of $\cos 12 ^{\circ} + \cos 84 ^{\circ} + \cos 156 ^{\circ} + \cos 132 ^{\circ}$ is

A. $\frac{1}{2}$

B. $1$
C. $-\frac{1}{2}$
D. $\frac{1}{8}$

Answer:

The answer is the option (c)
$\\ \cos 12+\cos 84+\cos 156+\cos 132= \left( \cos 132+\cos 12 \right) + \left( \cos 156+\cos 84 \right) \\\\ =2\cos \frac{132+12}{2}\cos \frac{132 - 12}{2}+2\cos \frac{156+84}{2}\cos \frac{156 - 84}{2} \\\\ =2\cos 72\cos 60+2\cos 120\cos 36 \\\\ =2\cos 72 \times \frac{1}{2}+2 \times \left( - \frac{1}{2} \right) \cos 36=\cos 72 - \cos 36 \\\\ =\frac{\sqrt {5} - 1}{4} - \frac{\sqrt {5}+1}{4}= - \frac{2}{4}= - \frac{1}{2} \\\\$

Question:47

If $\tan A=\frac{1}{2} , \tan B=\frac{1}{3} \\\\$ then $\tan (2A + B)$ is equal to
A. 1
B. 2
C. 3
D. 4

Answer:

The answer is the option (c).
$\\ \tan A=\frac{1}{2} \\\\ \tan B=\frac{1}{3} \\\\ \tan 2A=\frac{2\tan A}{1 - \tan ^{2}A}=\frac{4}{3} \\\\ \tan \left( 2A+B \right) =\frac{\tan 2A+\tan B}{1 - \tan 2A\tan B} \\\\ =\frac{\frac{4}{3}+\frac{1}{3}}{1 - \frac{4}{3} \times \frac{1}{3}}= \left( \frac{5}{3} \right) \times \left( \frac{9}{5} \right) =3 \\\\$

Question:48

The value of $\sin \frac{ \pi }{10}\sin \frac{13 \pi }{10}$ is
A. $\frac{1}{2}$
B. $-\frac{1}{2}$
C. $-\frac{1}{4}$
D. $1$

Answer:

The answer is the option (c).
$\\ \sin \frac{ \pi }{10}\sin \frac{13 \pi }{10}=\sin \frac{ \pi }{10}\sin \left( \pi +\frac{3 \pi }{10} \right) =\sin \frac{ \pi }{10} \left( - \sin \frac{3 \pi }{10} \right) \\\\ = - \sin 18\sin 54= - \left( \frac{\sqrt {5} - 1}{4} \right) \left( \frac{\sqrt {5}+1}{4} \right) \\\\ =\frac{5 - 1}{16}=\frac{4}{16}=\frac{1}{4} \\\\$
Hence, (c) is correct option.

Question:49

The value of $\sin 50 ^{\circ} - \sin 70 ^{\circ} + \sin 10 ^{\circ}$ is equal to
A. 1
B. 0
C.1/2
D. 2

Answer:

The answer is the option (b).

$\\\\ \sin 50 - \sin 70+\sin 10=2\cos \frac{50+70}{2}\sin \frac{50 - 70}{2}+\sin 10 \\\\ =2\cos 60\sin \left( - 10 \right) +\sin 10 \\\\ = - 2 \times \frac{1}{2} \times \sin 10+\sin 10=0 \\\\$

Question:50

If $\sin \theta + \cos \theta = 1$ , then the value of $\sin 2 \theta$ is equal to
A. 1
B. 1/2
C. 0
D. –1

Answer:

The answer is the option (c).
$\\\\ \sin \theta +\cos \theta =1 \\\\ \left( \sin \theta +\cos \theta \right) ^{2}=1 \\\\ \sin ^{2} \theta +\cos ^{2} \theta +2\sin \theta \cos \theta =1~~ \\\\ ~ 1+2\sin \theta \cos \theta =1 \\\\ 2\sin \theta \cos \theta =0 \\\\ \sin 2 \theta =0 \\\\$

Question:51

If $\alpha +\beta =\frac{\pi}{4}$ then the value of $(1+ \tan \alpha ) (1 + \tan \beta )$ is
A. 1
B. 2
C. –2
D. Not defined

Answer:

$\\ \tan \left( \alpha + \beta \right) =\tan \frac{ \pi }{4}=1 \\\\ \tan \left( \alpha + \beta \right) =\frac{\tan \alpha +\tan \beta }{1 - \tan \alpha \tan \beta }=1 \\\\ \tan \alpha +\tan \beta =1 - \tan \alpha \tan \beta \\\\ \tan \alpha +\tan \beta +\tan \alpha \tan \beta =1 \\\\ 1+\tan \alpha +\tan \beta +\tan \alpha \tan \beta =1+1 \\\\ \left( 1+\tan \alpha \right) \left( 1+\tan \beta \right) =2 \\\\$
Hence, correct option is (b).

Question:52

If $\sin \theta = - \frac{4}{5}$ and θ lies in third quadrant then the value of $\cos \frac{ \theta }{2}$ is
A. $\frac{1}{5}$
B. $\frac{-1}{\sqrt{10}}$
C. $\frac{-1}{\sqrt{5}}$
D. $\frac{1}{\sqrt{10}}$

Answer:

The answer is the option (c)
$\\ \sin \theta = - \frac{4}{5}, \theta \text{lies in third quadrant} \\\\$
$\\ \cos \theta = - \sqrt {1 - \left( - \frac{4}{5} \right) ^{2}}= - \frac{3}{5} \\\\ \cos \theta =2\cos ^{2}\frac{ \theta }{2} - 1 \\\\ - \frac{3}{5}=2\cos ^{2}\frac{ \theta }{2} - 1 \\\\ \cos ^{2}\frac{ \theta }{2}=\frac{1}{5} \\\\ \cos \frac{ \theta }{2}= - \frac{1}{\sqrt {5}}~ \\\\$
$\left[ As,~\frac{ \pi }{2}<\frac{ \theta }{2}<\frac{3 \pi }{4} \right]$
Hence, correct option is (c).

Question:53

Number of solutions of the equation $\tan x + sec x = 2 \cos x$ lying in the interval $[0, 2 \pi ]$ is
A. 0
B. 1
C. 2
D. 3

Answer:

The answer is the option (c).
$\\\\ \tan x+\sec x=2\cos x \\\\ \frac{\sin x+1}{\cos x}=2\cos x \\\\ 1+\sin x - 2\cos ^{2}x=0 \\\\ 1+\sin x - 2+2\sin ^{2}x=0 \\\\ 2\sin ^{2}x+\sin x - 1=0 \\\\$
Since the equation is a quadratic equation in $\sin x$ . So, there will be two solutions.
Hence, correct option is (c).

Question:54

The value of $\sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18}$ is given by
$\\A. \sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9}\\\\ B .1\\\\ C \cdot \cos \frac{\pi}{6}+\cos \frac{3 \pi}{7}\\\\ D \cdot \cos \frac{\pi}{9}+\sin \frac{\pi}{9}$

Answer:

The answer is the option (a).
$\\\\ \sin \frac{ \pi }{18}+\sin \frac{ \pi }{9}+\sin \frac{2 \pi }{9}+\sin \frac{5 \pi }{18}= \left( \sin \frac{ \pi }{18}+\sin \frac{5 \pi }{18} \right) + \left( \sin \frac{ \pi }{9}+\sin \frac{2 \pi }{9} \right) \\\\ =2\sin \frac{\frac{5 \pi }{18}+\frac{ \pi }{18}}{2}\cos \frac{\frac{5 \pi }{18} - \frac{ \pi }{18}}{2}+2\sin \frac{\frac{ \pi }{9}+\frac{2 \pi }{9}}{2}\cos \frac{\frac{2 \pi }{9} - \frac{ \pi }{9}}{2} \\\\ =2\sin \frac{ \pi }{6}\cos \frac{ \pi }{9}+2\sin \frac{ \pi }{6}\cos \frac{ \pi }{18}=2 \times \frac{1}{2}\cos \frac{ \pi }{9}+2 \times \frac{1}{2}\cos \frac{ \pi }{18} \\\\ =\cos \frac{ \pi }{9}+\cos \frac{ \pi }{18} \\\\ =\sin \left( \frac{ \pi }{2} - \frac{ \pi }{9} \right) +\sin \left( \frac{ \pi }{2} - \frac{ \pi }{18} \right) \\\\ =\sin \frac{4 \pi }{9}+\sin \frac{7 \pi }{18} \\\\$
Hence, correct option is (a).

Question:55

If A lies in the second quadrant and $3 \tan A + 4 = 0$ , then the value of $2 cot A- 5 \cos A + \sin A$ is equal to

A. $-\frac{53}{10}$
B. $\frac{23}{10}$
C. $\frac{37}{10}$
D. $\frac{7}{10}$

Answer:

The answer is the option (b).

$3\tan A+4=0$ [A lies in second quadrant]

$\tan A= - \frac{4}{3} \\\\$

$\cos A= - \frac{3}{5}$ [A lies in second quadrant]

$\\ \sin A=\frac{4}{5} \\\\ \cot A= - \frac{3}{4} \\\\ 2\cot A - 5\cos A+\sin A=2 \left( - \frac{3}{4} \right) - 5 \left( - \frac{3}{5} \right) +\frac{4}{5}= - \frac{3}{2}+3+\frac{4}{5}=\frac{23}{10} \\\\$

Hence, correct option is (b).

Question:56

The value of $\cos^2 48 ^{\circ} - \sin^2 12 ^{\circ}$ is

$\\A.\frac{\sqrt{5}+1}{8}\\\\ B.\frac{\sqrt{5}-1}{8}\\\\ C.\frac{\sqrt{5}+1}{5}\\\\ D.\frac{\sqrt{5}+1}{2 \sqrt{2}}\\\\$
$[Hint: Use\cos ^{2} A-\sin ^{2} B=\cos (A+B) \cos (A-B)]$

Answer:

The answer is the option (a).
$\\\\ \cos ^{2}48 - \sin ^{2}12=\cos \left( 48+12 \right) \cos \left( 48 - 12 \right) =\cos 60\cos 36=\frac{1}{2} \times \frac{\sqrt {5}+1}{4}=\frac{\sqrt {5}+1}{8} \\\\$

Hence, correct option is (a).

Question:57

If $\tan \alpha =\frac{1}{7} , \tan \beta =\frac{1}{3} \\\\$ then $\cos 2 \alpha$ is equal to
A. $\sin 2 \beta$
B. $\sin 4 \beta$
C. $\sin 2 \beta$
D. $\cos 2 \beta$

Answer:

The answer is the option (b).
$\\\\\\ \tan \alpha =\frac{1}{7} \\\\ \tan \beta =\frac{1}{3} \\\\ \cos 2 \alpha =\frac{1 - \tan ^{2} \alpha }{1+\tan ^{2} \alpha }=\frac{1 - \frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25} \\\\ \tan 2 \beta =\frac{2\tan \beta }{1 - \tan ^{2} \beta }=\frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}}=\frac{3}{4} \\\\$
$\\ \sin 4 \beta =\frac{2\tan 2 \beta }{1+\tan ^{2}2 \beta }=\frac{2 \times \frac{3}{4}}{1+ \left( \frac{3}{4} \right) ^{2}}=\frac{24}{25} \\\\ \cos 2 \alpha =\sin 4 \beta =\frac{24}{25} \\\\$
Hence, correct option is (b).

Question:58

If $\tan \theta =\frac{a}{b}$ then $b \cos 2 \theta + a \sin 2 \theta$ is equal to

$\\A. a \\\\ B. b\\\\ C. \frac{a}{b} \\\\ D. None$

Answer:

The answer is the option (b).
$\\\\ \tan \theta =\frac{a}{b} \\\\ b\cos 2 \theta +a\sin 2 \theta =b \left[ \frac{1 - \tan ^{2} \theta }{1+\tan ^{2} \theta } \right] +a \left[ \frac{2\tan \theta }{1+\tan ^{2} \theta } \right] \\\\ =b \left[ \frac{1 - \frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}} \right] +a \left[ \frac{2\frac{a}{b}}{1+\frac{a^{2}}{b^{2}}} \right] =b \left[ \frac{b^{2} - a^{2}}{b^{2}+a^{2}} \right] + \left[ \frac{\frac{2a^{2}}{b}}{\frac{b^{2}+a^{2}}{b^2}} \right] \\\\$
$\\ =\frac{b^{3} - a^{2}b}{b^{2}+a^{2}}+\frac{2a^{2}b}{b^{2}+a^{2}} \\\\ =\frac{b \left( b^{2}+a^{2} \right) }{b^{2}+a^{2}}=b \\\\$
Hence, correct option is (b).

Question:59

If for real values of x, $\cos \theta =x+\frac{1}{x}$ then
A. θ is an acute angle
B. θ is right angle
C. θ is an obtuse angle
D. No value of θ is possible

Answer:

The answer is the option (d).
$\\\\ \cos \theta =x+\frac{1}{x}=\frac{x^{2}+1}{x} \\\\ x^{2} - x\cos \theta +1=0 \\\\ \text{For real value of x, } \left( b \right) ^{2} - 4 \times a \times c \geq 0 \\\\$

$\\ \left( - \cos \theta \right) ^{2} - 4 \times 1 \times 1 \geq 0 \\\\ \cos ^{2} \theta \geq 4 \\\\ \cos \theta \geq \pm 2 \\\\$
Hence, correct option is (d).

Question:60

The value of $\frac{\sin 50}{\sin 130}$ is _______.

Answer:

$\\\frac{\sin 50}{\sin 130}=\frac{\sin 50}{\sin \left( 180 - 50 \right) }=\frac{\sin 50}{\sin 50}=1 \\\\$

Question:61

Fill in the blanks
If $k=\sin \left( \frac{ \pi }{18} \right) \sin \left( \frac{5 \pi }{18} \right) \sin \left( \frac{7 \pi }{18} \right) \\\\$ then the numerical value of k is

Answer:

$\\ k=\sin \left( \frac{ \pi }{18} \right) \sin \left( \frac{5 \pi }{18} \right) \sin \left( \frac{7 \pi }{18} \right) \\\\ k=\sin 10\sin 50\sin 70 \\\\ k=\sin 10\cos 40\cos 20 \\\\ k=\sin 10\frac{1}{2} \times \left[ 2\cos 40\cos 20 \right] =\sin 10\frac{1}{2} \left[ \cos 60+\cos 20 \right] \\\\ k=\frac{1}{2}\sin 10 \left[ \frac{1}{2}+\cos 20 \right] \\\\ k=\frac{1}{4}\sin 10+\frac{1}{2}\sin 10\cos 20 \\\\ k=\frac{1}{4}\sin 10+\frac{1}{4} \left[ \sin 30 - \sin 10 \right] \\\\ k=\frac{1}{4}\sin 30=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8} \\\\$

Question:62

Fill in the blanks

If $\tan A=\frac{1 - \cos B}{\sin B}$ then $\tan 2A =$ ......

Answer:

$\\ \tan A=\frac{1 - \cos B}{\sin B} \\\\ \tan 2A=\frac{2\tan A}{1+\tan ^{2}A}=\frac{\frac{2 \left( 1 - \cos B \right) }{\sin B}}{1+\frac{ \left( 1 - \cos B \right) ^{2}}{\sin ^{2}B}} \\\\ =\frac{2 \left( \frac{2\sin ^{2}\frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}} \right) }{1 - \left( \frac{2\sin ^{2}\frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}~} \right) ^{2}} \\\\ =\frac{2 \left( \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} \right) }{1 - \left( \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} \right) ^{2}}=\frac{2\tan \frac{B}{2}}{1 - \tan ^{2}\frac{B}{2}}=\tan B \\\\$

Question:63

Fill in the blanks
If $\sin x + \cos x = a$ , then
(i) $\sin^6 x + \cos^6x = \_\_\_\_\_\_\_$
(ii) $\vert \sin x - \cos x \vert = \_\_\_\_\_\_.$

Answer:

$\\Given\: \: that \: \: \sin x+\cos x=a \\\\ \text{On squaring both sides}, \left( \sin x+\cos x \right) ^{2}=a^{2} \\\\ 1+2\sin x\cos x=a^{2} \\\\ \sin x\cos x=\frac{a^{2} - 1}{2} \\\\ \sin ^{6}x+\cos ^{6}x= \left( \sin ^{2}x \right) ^{3}+ \left( \cos ^{2}x \right) ^{3} \\\\ = \left( \sin ^{2}x+\cos ^{2}x \right) ^{3} - 3\sin ^{2}x\cos ^{2}x \left( \sin ^{2}x+\cos ^{2}x \right) \\\\ =1 - 3 \left( \frac{a^{2} - 1}{2} \right) ^{2}=1 - \frac{3 \left( a^{2} - 1 \right) ^{2}}{4}=\frac{1}{4} \left[ 4 - 3 \left( a^{2} - 1 \right) ^{2} \right] \\\\ \vert \sin x - \cos x \vert ^{2}=\sin ^{2}x+\cos ^{2}x - 2\sin x\cos x \\\\ =1 - 2 \left( \frac{a^{2} - 1}{2} \right) =1 - a^{2}+1=2 - a^{2} \\\\ \vert \sin x - \cos x \vert =\sqrt {2 - a^{2}} \\\\$

Question:64

Fill in the blanks

In a triangle ABC with $\angle C = 90 ^{\circ}$ the equation whose roots are tan A and tan B is ________.
[Hint: $A + B = 90 ^{\circ} \Rightarrow \tan A \tan B = 1 and \tan A + \tan B = \frac{2}{ \sin 2A }$ ]

Answer:

$\\\\ x^{2} - \left( \tan A+\tan B \right) x+\tan A\tan B=0 \\\\ \tan \left( A+B \right) =\tan 90 \\\\ \frac{\tan A+\tan B}{1 - \tan A\tan B}=\frac{1}{0} \\\\ 1 - \tan A\tan B=0 \\\\ \tan A\tan B=1 \\\\ \tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}=\frac{\sin \left( A+B \right) }{\cos A\cos B}=\frac{1}{\cos A\sin A} \\\\ \tan A+\tan B=\frac{2}{2\sin A\cos A}=\frac{2}{\sin 2A}~ \\\\ x^{2} - \left( \frac{2}{\sin 2A} \right) x+1=0$

Question:65

Fill in the blanks
$3(\sin x - \cos x)^4 + 6 (\sin x + \cos x)^2 + 4(\sin^6 x + \cos^6 x) =$

Answer:

$\\\\\\ 3 \left( \sin x - \cos x \right) ^{4}+6 \left( \sin x+\cos x \right) ^{2}+4 \left( \sin ^{6}x+\cos ^{6}x \right) \\\\ =3 \left( \sin ^{2}x+\cos ^{2}x - 2\sin x\cos x \right) ^{2}+6 \left( \sin ^{2}x+\cos ^{2}x+2\sin x\cos x \right) +4 \left[ \left( \sin ^{2}x \right) ^{3}+ \left( \cos ^{2}x \right) ^{3} \right] \\\\ =3 \left( 1 - 2\sin x\cos x \right) ^{2}+6+12\sin x\cos x+4 \left[ \left( \sin ^{2}x+\cos ^{2}x \right) ^{3} - 3\sin ^{2}x\cos ^{2}x \left( \sin ^{2}x+\cos ^{2}x \right) \right] \\\\ =3 \left( 1+4\sin ^{2}x\cos ^{2}x - 4\sin x\cos x \right) +6+12\sin x\cos x+4 - 12\sin ^{2}x\cos ^{2}x \\\\ =3+12\sin ^{2}x\cos ^{2}x - 12\sin x\cos x+6+12\sin x\cos x+4 - 12\sin ^{2}x\cos ^{2}x \\\\ =3+6+4=13 \\\\$

Question:66

Fill in the blanks
Given x > 0, the values of $f \left( x \right) = - 3\cos \sqrt {3+x+x^{2}}$ lie in the interval _______.

Answer:

$\\Given\: \: that \: \: f \left( x \right) = - 3\cos \sqrt {3+x+x^{2}} \\\\ Putting, y=\sqrt {3+x+x^{2}} \\\\ f \left( x \right) = - 3\cos y \\\\ - 1 \leq \cos y \leq 1 \\\\ \vspace{\baselineskip} - 3 \leq - 3\cos\sqrt {3+x+x^{2}} \leq 3 \\\\ \text{Hence the value is in } [ - 3,3] \\\\$

Question:67

Fill in the blanks
The maximum distance of a point on the graph of the function $y=\sqrt {3}\sin x+\cos x$ from x-axis is _____.

Answer:

$\\\\\\ y=\sqrt {3}\sin x+\cos x \ldots \ldots . (i) \\\\$
The maximum distance from a point on the graph of equation (i) from x - axis
$\\ \sqrt { \left( \sqrt {3} \right) ^{2}+ \left( 1 \right) ^{2}}=\sqrt {3+1}=2 \\\\$

Question:68

True and False

If $\tan A=\frac{1 - \cos B}{\sin B}$ then $\tan 2A = \tan B$

Answer:

$\\ \tan A=\frac{1 - \cos B}{\sin B}=\frac{2\sin ^{2}\frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}}=\tan \frac{B}{2} \\\\ \tan 2A=\tan B \\\\$

Hence, the statement is true.

Question:69

True and False
The equality $\sin A + \sin 2A + \sin 3A = 3$ holds for some real value of A.

Answer:

Given that $\sin A+\sin 2A+\sin 3A=3 \\\\$
Since the maximum value of sin A is 1 but for sin 2A and sin 3A it is not equal to 1. So, it is not possible.
Hence, the statement is ’false’.

Question:70

True and False
$\sin 10 ^{\circ}$ is greater than $\cos 10 ^{\circ}$

Answer:

$\\If\ \sin 10>\cos 10 \\\\ Then, \sin 10>\cos \left( 90 - 80 \right) \\\\ \sin 10>\sin 80$
which is not possible because value of sine is in increasing order
Hence, the statement is ‘false’

Question:71

True and False
$\cos \frac{2 \pi }{15}\cos \frac{4 \pi }{15}\cos \frac{8 \pi }{15}\cos \frac{16 \pi }{15} = \frac{1}{16}$

Answer:

$\\\cos \frac{2 \pi }{15}\cos \frac{4 \pi }{15}\cos \frac{8 \pi }{15}\cos \frac{16 \pi }{15} \\\\ =\cos 24\cos 48\cos 96\cos 192 \\\\ =\frac{1}{16\sin 24} \left( 2\sin 24\cos 24 \right) \left( 2\cos 48 \right) \left( 2\cos 96 \right) \left( 2\cos 192 \right) \\\\ =\frac{1}{16\sin 24} \left( 2\sin 48\cos 48 \right) \left( 2\cos 96 \right) \left( 2\cos 192 \right) \\\\ =\frac{1}{16\sin 24} \left( 2\cos 96\sin 96 \right) \left( 2\cos 192 \right) =\frac{2\sin 192\cos 192}{16\sin 24} \\\\ =\frac{\sin 384}{16\sin 24}=\frac{\sin \left( 360+24 \right) }{16\sin 24} \\\\ =\frac{\sin 24}{16\sin 24}=\frac{1}{16} \\\\$
Hence, the statement is ‘true’.

Question:72

True and False
One value of θ which satisfies the equation $\sin^4 \theta - 2\sin^2 \theta - 1$ lies between 0 and 2π

Answer:

Given equation is $\sin ^{4} \theta - 2\sin ^{2} \theta - 1=0 \\\\$

$\\ \sin ^{2} \theta =\frac{ - \left( - 2 \right) \pm \sqrt { \left( - 2 \right) ^{2} - 4 \times 1 \times \left( - 1 \right) }}{2 \times 1}=\frac{2 \pm \sqrt {4+4}}{2} \\\\ =\frac{2 \pm \sqrt {8}}{2}=\frac{2 \pm 2\sqrt {2}}{2}=1 \pm \sqrt {2} \\\\$
$- 1 \leq \sin \theta \leq 1~ and \sin ^{2} \theta \leq 1~ but \sin ^{2} \theta =1 \pm \sqrt {2}$
which is not possible
Hence, the given statement is ‘false’

Question:73

True and False
If $cosec x = 1 + cot x$ then $x=2n \pi ,2n\pi+\frac{\pi}{2}$

Answer:
$\\cosec x=1+\cot x~ \\\\ x=2n \pi ,2n \pi +\frac{ \pi }{2} \\\\ \frac{1}{\sin x}=1+\frac{\cos x}{\sin x} \\\\ \sin x+\cos x=1 \\\\$
$\\\frac{1}{\sqrt {2}}\sin x+\frac{1}{\sqrt {2}}\cos x=\frac{1}{\sqrt {2}} \\\\ \cos \left( x - \frac{ \pi }{4} \right) =\cos \frac{ \pi }{4} \\\\ x=2n \pi +\frac{ \pi }{4}+\frac{ \pi }{4}=2n \pi +\frac{ \pi }{2} \\\\ Or, x=2n \pi + \frac{ \pi }{4} - \frac{ \pi }{4}= 2n \pi \\\\$
Hence, the given statement is ‘true’

Question:74

True and False

$\tan \theta + \tan 2 \theta + \sqrt{3} \tan \theta \tan 2 \theta = \sqrt{3}$ then $\theta =\frac{n \pi }{3}+\frac{ \pi }{9}$

Answer:

$\\ \tan \theta +\tan 2 \theta = - \sqrt {3}\tan \theta \tan 2 \theta +\sqrt {3} \\\\ \tan \theta +\tan 2 \theta =\sqrt {3} \left( 1 - \tan \theta \tan 2 \theta \right) \\\\ \frac{\tan \theta +\tan 2 \theta }{1 - \tan \theta \tan 2 \theta }=\sqrt {3} \\\\ \tan 3 \theta =\sqrt {3} \\\\ \tan 3 \theta =\tan \frac{ \pi }{3} \\\\ 3 \theta =n \pi +\frac{ \pi }{3} \\\\ \theta =\frac{n \pi }{3}+\frac{ \pi }{9} \\\\$
Hence, the given statement is ‘true’ .

Question:75

True and False
If $\tan( \pi \cos \theta ) = cot ( \pi \sin \theta )$ , then $\cos \left( \theta - \pi /4 \right) = \pm \frac{1}{2\sqrt {2}}$

Answer:

$\\ \tan \left( \pi \cos \theta \right) =\cot \left( \pi \sin \theta \right) \\\\ \tan \left( \pi \cos \theta \right) =\tan \left( \frac{ \pi }{2} - \pi \sin \theta \right) \\\\ \pi \cos \theta =\frac{ \pi }{2} - \pi \sin \theta \\\\ \cos \theta +\sin \theta =\frac{1}{2} \\\\ ~\cos \frac{ \pi }{4}~\cos \theta +\sin \frac{ \pi }{4}\sin \theta =\frac{1}{2} \\\\ \vspace{\baselineskip} \cos \left( \theta - \pi /4 \right) = \pm \frac{1}{2\sqrt {2}} \\\\$
Hence, the given statement is ‘true’

Question:76

In the following match each item given under the column $C_1$ to its correct answer given under the column $C_2$ :

$\text { (a) } \sin (\mathrm{x}+\mathrm{y}) \sin (\mathrm{x}-\mathrm{y})$	$\text { (i) } \cos ^{2} \mathrm{x}-\sin ^{2} \mathrm{y}$
$\text { (b) } \cos (\mathrm{x}+\mathrm{y}) \cos (\mathrm{x}-\mathrm{y})$	$\text { (ii) } \frac{1-\tan \theta}{1+\tan \theta}$
$\text { (c) } \cot \left(\frac{\pi}{4}+\theta\right)$	$\text { (iii) } \frac{1+\tan \theta}{1-\tan \theta}$
$\text { (d) } \tan \left(\frac{\pi}{4}+\theta\right)$	$\text { (iv) } \sin ^{2} \mathrm{x}-\sin ^{2} \mathrm{y}$

Answer:

$\\ \sin \left( x+y \right) \sin \left( x - y \right) =\sin ^{2}x - \sin ^{2}y \\\\ \cos \left( x+y \right) \cos \left( x - y \right) =\cos ^{2}x - \cos ^{2}y \\\\ \cot \left( \frac{ \pi }{4}+ \theta \right) =\frac{\cot \frac{ \pi }{4}\cot \theta - 1}{\cot \theta +\cot \frac{ \pi }{4}}=\frac{\cot \theta - 1}{\cot \theta +1}=\frac{1 - \tan \theta }{1+\tan \theta } \\\\ \tan \left( \frac{ \pi }{4}+ \theta \right) =\frac{\tan \frac{ \pi }{4}\tan \theta - 1}{\tan \theta +\tan \frac{ \pi }{4}}=\frac{1+\tan \theta }{1 - \tan \theta } \\\\$
Thus, (a) - (iv) , (b) - (i), (c) -(ii), (d) - (iii)

Important Notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Trigonometry is an ancient concept which in ancient times was used to solve problems relating to triangles and geometry, but it has extended its use to various different fields in the present times including varied areas of studies. It was derived from Greek words meaning, “measuring the sides of a triangle” which has widened its scope to much more than the original meaning. It is basically used to measure length, height and angles of different triangles with its reach in real-life practical situations. NCERT Exemplar solutions for Class 11 Maths chapter 3 extends studying trigonometric ratios to any angle regarding or concerning radian measure and interpreting and representing it as a trigonometric ratio with the help of diagrams for a better understanding.

Students can make use of NCERT Exemplar Class 11 Maths solutions chapter 3 pdf download for further learning.

Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3

The topics covered in the chapter are as follows:

3.1 Introduction
3.2 Angles
3.2.1 Degree measure
3.2.2 Radian measure
3.2.3 Relation between radian and real numbers
3.2.4 Relation between degree and radian
3.3 Trigonometric Functions
3.3.1Sign of trigonometric functions
3.3.2 Domain and range of trigonometric functions
3.4 Trigonometric Functions of Sum and Difference of Two Angles
3.5 Trigonometric Equations

What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?

The students will learn a variety of concepts from NCERT Exemplar solutions for Class 11 Maths chapter 3 which has a wide range of application in different fields such as engineering, sound engineers, architects, astronauts, surveyors, physicist, and much more for future references. NCERT Exemplar Class 11 Maths solutions chapter 3 is also given importance since it has various applications in real life and could be connected to routine activities that happen around us. It is even used in the gaming industry, IT sector, construction of bridges, buildings, mountains, the inclination of floors, roofs, marine biology, criminal investigations, wide use in physics for derivations and explanations, and much more. The uses of trigonometry in these many fields justify its use for inexhaustible purposes and its importance for students belonging or deciding to enter any field or subject in the future.

NCERT Solutions for Class 11 Mathematics Chapters

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

· NCERT Exemplar Class 11 Maths chapter 3 solutions give explanation, and interpretation of different trigonometric functions for sin x, cos x, sec x, cot x, cosec x, and tan x along with values of trigonometric ratios for 0º, 30º, 45º, 60º, 90º, 180º, 270º and 360º along with the sign convention of these trigonometric functions.

· Class 11 Maths NCERT Exemplar solutions chapter 3 also covers the range and domain of trigonometric functions with the help of diagrammatic representation of the same. This chapter also extends to the trigonometric functions of sum and difference of two angles with a variety of questions and illustrations to be done for the same.

· NCERT Exemplar class 11 Maths solutions chapter 3 concludes with insight on trigonometric equations that involve equations containing trigonometric functions of any variable.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Physics Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Who can use these solutions of the NCERT chapter?

Those who are prepping for their board exams and for those who are planning to appear in their JEE Main exam or any other engineering entrance exam.

2. What all topics are covered in this chapter?

This chapter covers everything related to trigonometric functions, their properties, angles, trigonometric equations, etc.

3. Who has prepared these NCERT exemplar Class 11 maths solutions chapter 3?

These solutions are prepared by other experienced maths teachers and team so as to include every detail of the solution with no mistake.

4. How to make use of these solutions?

Students can practice the questions, and while doing so, these Class 11 Maths NCERT exemplar solutions chapter 3 will act as a reference while studying.

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NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3

Important Notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3

What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Solutions

Frequently Asked Question (FAQs)

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