NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Edited By Komal Miglani | Updated on Mar 30, 2025 05:50 PM IST

While playing cricket, have you ever wondered how we calculate the trajectory or range of the ball or has your curious mind ever thought of constructing a building or a monument? The answer to all these is Trigonometry! Trigonometric functions is a fundamental and important chapter of NCERT. This chapter helps students gain ideas about angles and their measurements in terms of degree or radians, the domain and range of trigonometric functions and their related graphs. It also helps students learn necessary trigonometric identities and formulas and allows students to dive deep into more advanced concepts like periodicity and transformations and properties of sine, cosine and other trigonometric ratios. The students preparing for their competitive exams are advised to practice their worksheets and exercises on a regular basis as the questions asked are of the same pattern as provided in the NCERT Solutions.

This Story also Contains
  1. NCERT Exemplar Class 11 Maths Solutions Chapter 3
  2. Importance of Solving NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions
  3. Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3
  4. NCERT Solutions for Class 11 Mathematics Chapters
  5. NCERT Exemplar Class 11 Mathematics Chapters
  6. NCERT solutions of class 11 - Subject-wise
  7. NCERT Notes of class 11 - Subject Wise
  8. NCERT Books and NCERT Syllabus

NCERT Exemplar Class 11 Maths Solutions Chapter 3

Class 11 Maths Chapter 3 Exemplar Solutions Exercise: 3.3
Page number: 52-60
Total questions: 76

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

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

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 dividendo]

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 $\sin 4 A=4 \sin A \cos ^3 A-4 \cos A \sin ^3 A$.

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^2-n^2=4 \sin \theta \tan \theta
$

$\left[\right.$ Hint: $m+n=2 \tan \theta, m-n=2 \sin \theta$, then use $\left.\left[m^2-n^2\right]=(m+n)(m-n)\right]$

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: $\left.(\cos \alpha+\cos \beta)^2-(\sin \alpha+\sin \beta)^2=0\right]$

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 $2 \sin ^2 \theta=3 \cos \theta$, where $0 \leq \theta \leq 2 \pi$, then find the value of $\theta$.

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 $\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta)=1-2 a^2-2 b^2$

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 $a \cos 2 \theta+b \sin 2 \theta=c$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan \alpha+\tan \beta=2 b /(a+c)$ [Hint: Use the identities $\cos 2 \theta=\left(\left(1-\tan ^2 \theta\right) /\left(1+\tan ^2 \theta\right)\right.$ and $\left.\sin 2 \theta=2 \tan \theta /\left(1+\tan ^2 \theta\right)\right]$

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 ^4(\pi / 8)+\cos ^4(3 \pi / 8)+\cos ^4(5 \pi / 8)+\cos ^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 ^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 $\sin ^2 \theta+\operatorname{cosec}^2 \theta$ is equal to
A. 1
B. 4
C. 2
D. None of these

Answer:

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

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

The answer is the option (c).

Question:31

If $f(x)=\cos ^2 x+\sec ^2 x$, 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:

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

The answer is the option (d)

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 $\left(1-\tan ^2 15^{\circ}\right) /\left(1+\tan ^2 15^{\circ}\right)$ 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: 1radian $=180^{\circ} \pi=57^{\circ} 30^{\prime}$approx.]

Answer:
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 $\alpha+\beta_{\text {is equal to }}$
$\\A. \frac{\pi}{2}\\\\ B.\frac{\pi}{3}\\\\ C.\frac{\pi}{c} \\\\ D.\frac{\pi}{4}$

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:
$\\Let\ y=3\cos x+4\sin x+8 \\\\ y - 8= 3\cos x+4\sin x \\\\ \text{Minimum value of } y - 8= - \sqrt { \left( 3 \right) ^{2}+ \left( 4 \right) ^{2}}= - 5 \\\\ y=8 - 5=3 \\\\$
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).
$\\ \cot \left( \frac{ \pi }{4}+ \theta \right) \cot \left( \frac{ \pi }{4} - \theta \right) =\frac{\cot \frac{ \pi }{4}\cot \theta - 1}{\cot \theta +\cot \frac{ \pi }{4}} \times \frac{\cot \frac{ \pi }{4}\cot \theta +1}{\cot \theta - \cot \frac{ \pi }{4}} \\\\ =\frac{\cot \theta - 1}{\cot \theta +1} \times \frac{\cot \theta +1}{\cot \theta - 1}=1 \\\\$

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:
$\\ \sin \theta = - \frac{4}{5}, \theta \text{lies in third quadrant} \\\\$
$\\ \cos \theta = - \sqrt {1 - \left( - \frac{4}{5} \right) ^{2}}= - \frac{3}{5} \\\\ \cos \theta =2\cos ^{2}\frac{ \theta }{2} - 1 \\\\ - \frac{3}{5}=2\cos ^{2}\frac{ \theta }{2} - 1 \\\\ \cos ^{2}\frac{ \theta }{2}=\frac{1}{5} \\\\ \cos \frac{ \theta }{2}= - \frac{1}{\sqrt {5}}~ \\\\$
$\left[ As,~\frac{ \pi }{2}<\frac{ \theta }{2}<\frac{3 \pi }{4} \right]$
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:
$\\\\ \tan x+\sec x=2\cos x \\\\ \frac{\sin x+1}{\cos x}=2\cos x \\\\ 1+\sin x - 2\cos ^{2}x=0 \\\\ 1+\sin x - 2+2\sin ^{2}x=0 \\\\ 2\sin ^{2}x+\sin x - 1=0 \\\\$
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

$
\begin{aligned}
& C \cdot \cos \frac{\pi}{6}+\cos \frac{3 \pi}{7} \\
& D \cdot \cos \frac{\pi}{9}+\sin \frac{\pi}{9}
\end{aligned}
$

Answer:


$\\$$\\ \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:

$3\tan A+4=0$ [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, the 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:

$\\\\ \cos ^{2}48 - \sin ^{2}12=\cos \left( 48+12 \right) \cos \left( 48 - 12 \right) =\cos 60\cos 36=\frac{1}{2} \times \frac{\sqrt {5}+1}{4}=\frac{\sqrt {5}+1}{8} \\\\$
Hence, the 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:

$\\\\\\ \tan \alpha =\frac{1}{7} \\\\ \tan \beta =\frac{1}{3} \\\\ \cos 2 \alpha =\frac{1 - \tan ^{2} \alpha }{1+\tan ^{2} \alpha }=\frac{1 - \frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25} \\\\ \tan 2 \beta =\frac{2\tan \beta }{1 - \tan ^{2} \beta }=\frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}}=\frac{3}{4} \\\\$
$\\ \sin 4 \beta =\frac{2\tan 2 \beta }{1+\tan ^{2}2 \beta }=\frac{2 \times \frac{3}{4}}{1+ \left( \frac{3}{4} \right) ^{2}}=\frac{24}{25} \\\\ \cos 2 \alpha =\sin 4 \beta =\frac{24}{25} \\\\$
Hence, the 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:
$\\\\ \tan \theta =\frac{a}{b} \\\\ b\cos 2 \theta +a\sin 2 \theta =b \left[ \frac{1 - \tan ^{2} \theta }{1+\tan ^{2} \theta } \right] +a \left[ \frac{2\tan \theta }{1+\tan ^{2} \theta } \right] \\\\ =b \left[ \frac{1 - \frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}} \right] +a \left[ \frac{2\frac{a}{b}}{1+\frac{a^{2}}{b^{2}}} \right] =b \left[ \frac{b^{2} - a^{2}}{b^{2}+a^{2}} \right] + \left[ \frac{\frac{2a^{2}}{b}}{\frac{b^{2}+a^{2}}{b^2}} \right] \\\\$
$\\ =\frac{b^{3} - a^{2}b}{b^{2}+a^{2}}+\frac{2a^{2}b}{b^{2}+a^{2}} \\\\ =\frac{b \left( b^{2}+a^{2} \right) }{b^{2}+a^{2}}=b \\\\$
Hence, the 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 $

$-3 \leq -3\cos\sqrt{3 + x + x^{2}} \leq 3$

$ \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 the 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 the 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:

$\\ \tan \left( \pi \cos \theta \right) =\cot \left( \pi \sin \theta \right) \\$

$\\ \tan \left( \pi \cos \theta \right) =\tan \left( \frac{ \pi }{2} - \pi \sin \theta \right) \\$

$\\ \pi \cos \theta =\frac{ \pi }{2} - \pi \sin \theta \\$

$\\ \cos \theta +\sin \theta =\frac{1}{2} \\$

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

$\\ \cos \left( \theta - \pi /4 \right) = \pm \frac{1}{2\sqrt {2}} \\\\$


Hence, the given statement is ‘true’

Question:76

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

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)

Importance of Solving NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

  • NCERT exemplar for class 11 maths chapter 3 Trigonometric Functions is designed to give the students step-by-step solutions for a particular question.
  • Students can study strategically at their own pace after accessing Class 11 Maths NCERT exemplar chapter 3. This will boost their confidence to attempt other questions from this chapter.
  • These NCERT exemplar cover all the important topics and concepts so that students can be well-prepared for various exams.
  • By solving these NCERT trigonometry problems students will get to know about all the real-life applications of trigonometry.

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
  • 3.3.1 Sign 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

NCERT Solutions for Class 11 Mathematics Chapters

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NCERT Exemplar Class 11 Mathematics Chapters

Students can use following links to study NCERT exemplar solutions chapter wise.

NCERT solutions of class 11 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Notes of class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of class 11 :

NCERT Books and NCERT Syllabus

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NCERT Exemplar Class 11 Solutions

Given below are the subject-wise exemplar solutions of class 11 NCERT:



Frequently Asked Questions (FAQs)

1. What are the important topics in the NCERT Exemplar Class 11 Maths Chapter 3?

Trigonometric Functions deal with the angles in degrees and radians, trigonometric functions (sine, cosine, tangent, etc.), graph, domain, and range. It deals with the periodicity of functions and their transformations. Students also learn symmetry and if functions are even or odd. The chapter also deals with concepts like symmetry and even/odd functions. Solving NCERT exemplar problems helps in better comprehension of such concepts and preparation for the board exams.

2. How to solve NCERT Exemplar Class 11 Maths Chapter 3 problems easily?

Trigonometric functions should be handled confidently, beginning from small things such as their range, domain, and their graphs. Begin with simple problems so that you are right in your basics. Then, go and try tough problems. Attempt simple graph-based questions by representing functions in terms of graphs. Try to solve the examples first and then try to attempt the exercises. Practice regularly for speed and accuracy. Regular revision and practice will help you gain a good concept and try the NCERT Exemplar questions easily.

3. What are the real-life applications of trigonometric functions in Class 11?

Architecture and Building Construction – Used to compute heights, angles, and distances when designing buildings, monuments, and bridges.
Navigation and GPS – Supports determination of positions, distance measurement, and direction-finding on maps and GPS.
Astronomy – Used to find the distance of stars, planets, and where the Earth is in the universe.
Physics and engineering – Critical in the case of topics such as mechanics, wave motion, and electrical engineering, to teach one about vibration and forces.
Aviation and Marine Science Helps in calculating flight paths, shipping paths, and altitudes.
Sound and Music – Applied in the study of sound waves, frequency, and resonance of musical instruments.
Medical Imaging – Applied in CT scans, MRI, and ultrasound to generate high-resolution images of the human body.

4. Why is Chapter 3 Trigonometric Functions important in Class 11 Maths?

Foundation for Advanced Mathematics – Trigonometric functions are very significant in advanced-level calculus, coordinate geometry, and algebra.
Applications in Real Life – Applied in physics, engineering, astronomy, architecture, and navigation to find angles, distances, and heights.
Extremely Useful for Competitive Exams – Useful for JEE, NEET, and other entrance exams, as the majority of the questions are trigonometry-based.
Graphical Conceptualization – Enhances comprehension of periodic functions, transformations, and their applications in real life.
Key Identities and Formulas – Provides key formulas such as sin²θ + cos²θ = 1, which are frequently applied in equation solving

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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