NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

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)