Question 11 Mark
Prove that: ${\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right) = \frac{x}{2}$, $x \in \left( {0,\frac{\pi }{4}} \right)$
AnswerWe know that $1 + \sin x = {\cos ^2}\frac{x}{2} + {\sin ^2}\frac{x}{2} + 2\cos \frac{x}{2}\sin \frac{x}{2}$ $= {\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)^2}$
Again, $1 - \sin x = {\cos ^2}\frac{x}{2} + {\sin ^2}\frac{x}{2} - 2\cos \frac{x}{2}\sin \frac{x}{2}$
$= {\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2}$
$= {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right)$
$ = {\cot ^{ - 1}}\left[ {\frac{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) + \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}}{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) - \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}}} \right]$
$ = {\cot ^{ - 1}}\left( {\frac{{2\cos \frac{x}{2}}}{{2\sin \frac{x}{2}}}} \right)$
$= {\cot ^{ - 1}}\cot \frac{x}{2} = \frac{x}{2}$
View full question & answer→Question 21 Mark
Prove that: $\tan ^ { - 1 } \sqrt { x } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) , x \in ( 0,1 ).$
AnswerWe have to prove, $\tan ^ { - 1 } \sqrt { x } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) , x \in ( 0,1 ).$ Let ,$\tan ^ { - 1 } \sqrt { x }=\theta$ ....(i)
$\Rightarrow tan\theta=\sqrt{x}$
Let R.H.S = $= \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right)$$= \frac { 1 } { 2 } \cos ^ { - 1 } \left[ \frac { 1 - ( \sqrt { x } ) ^ { 2 } } { 1 + ( \sqrt { x } ) ^ { 2 } } \right]$
R.H.S = $\frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } \right)$ [from equation (i) ]
$= \frac { 1 } { 2 } \cos ^ { - 1 } ( \cos 2 \theta )$$\left[ \because \frac { 1 - \tan ^ { 2 } A } { 1 + \tan ^ { 2 } A } = \cos 2 A \right]$
$= \frac { 1 } { 2 } ( 2 \theta ) = \theta$$\left[ \because \cos ^ { - 1 } ( \cos \theta ) = \theta ; \forall \theta \in [ 0 , \pi ] \right]$
$= \tan ^ { - 1 } \sqrt { x }$$\left[ \because \theta = \tan ^ { - 1 } \sqrt { x } \right]$
= L.H.S Hence proved.
View full question & answer→Question 31 Mark
Prove that: ${\tan ^{ - 1}}\frac{{63}}{{16}} = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$
AnswerLet ${\sin ^{ - 1}}\frac{5}{{13}} = \theta$ so that $\sin \theta = \frac{5}{{13}}$
$\therefore \cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \frac{{25}}{{169}}}$$= \sqrt {\frac{{144}}{{169}}} = \frac{{12}}{{13}}$
$\therefore \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{5}{{12}}$
Again, Let ${\cos ^{ - 1}}\frac{3}{5} = \phi$ so that $\cos \phi = \frac{3}{5}$
$\therefore \sin \phi = \sqrt {1 - {{\cos }^2}\phi } = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$\therefore \tan \phi = \frac{{\sin \phi }}{{\cos \phi }} = \frac{4}{3}$
Since $\tan \left( {\theta + \phi } \right) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \times \frac{4}{3}}}$
$= \frac{{\frac{{21}}{{12}}}}{{\frac{4}{9}}} = \frac{{63}}{{16}}$
$\Rightarrow \theta + \phi = {\tan ^{ - 1}}\frac{{63}}{{16}}$
$\Rightarrow {\tan ^{ - 1}}\frac{{63}}{{16}} = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$
View full question & answer→Question 41 Mark
Prove that $\cos ^ { - 1 } \left( \frac { 12 } { 13 } \right) + \sin ^ { - 1 } \left( \frac { 3 } { 5 } \right) = \sin ^ { - 1 } \left( \frac { 56 } { 65 } \right).$
Answer$\text{LHS} = \cos^{-1} (\frac{12}{13}) + \sin^{-1} (\frac{3}{5})$
$= \sin^{-1}\sqrt{1-\left(\frac{12}{13}\right)^{2}} + \sin^{-1} \frac{3}{5}$
$= \sin^{-1} \sqrt{1-\frac{144}{169}} + \sin^{-1}\frac{3}{5}$
$= \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{3}{5}$
$= \sin^{-1} \left[\frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right]$
$= \sin^{-1} \left[\frac{5}{13} \sqrt{1-\frac{9}{25}}+\frac{3}{5} \sqrt{1-\frac{25}{169}}\right]$
$= \sin^{-1} \left[\frac{5}{13} \times \frac{4}{5}+\frac{3}{5} \times \frac{12}{13}\right]= \sin^{-1 }\left[\frac{20}{65}+\frac{36}{65}\right]$
$= \sin^{-1} \left[\frac{56}{65}\right] = \text{RHS}$
View full question & answer→Question 51 Mark
Prove that $ \cos ^ { - 1 } \left( \frac { 4 } { 5 } \right) + \cos ^ { - 1 } \left( \frac { 12 } { 13 } \right) = \cos ^ { - 1 } \left( \frac { 33 } { 65 } \right).$
AnswerHere we need to prove that $ \cos ^ { - 1 } \left( \frac { 4 } { 5 } \right) + \cos ^ { - 1 } \left( \frac { 12 } { 13 } \right) = \cos ^ { - 1 } \left( \frac { 33 } { 65 } \right)$
We consider that $ {\cos ^{ - 1}}\frac{4}{5} = x$ and $ \cos ^ { - 1 } \left( \frac { 12 } { 13 } \right) = y : \forall x , y \in [ 0 , \pi ]$
$ \Rightarrow \quad \cos x = \frac { 4 } { 5 }$ and $ \cos y = \frac { 12 } { 13 }$ .....(i)
$ \Rightarrow \quad \sin x = \sqrt { 1 - \cos ^ { 2 } x }$ and $ \sin y = \sqrt { 1 - \cos ^ { 2 } y }$ [taking positive sign as x, y,$ \in [ 0 , \pi ]$]
$ \Rightarrow \quad \sin x = \sqrt { 1 - \left( \frac { 4 } { 5 } \right) ^ { 2 } }$ and $ \sin y = \sqrt { 1 - \left( \frac { 12 } { 13 } \right) ^ { 2 } }$
$ \Rightarrow \quad \sin x = \sqrt { 1 - \frac { 16 } { 25 } }$ and $ \sin y = \sqrt { 1 - \frac { 144 } { 169 } }$
$ \Rightarrow \quad \sin x = \sqrt { \frac { 25 - 16 } { 25 } } = \sqrt { \frac { 9 } { 25 } }$
and $ \sin y = \sqrt { \frac { 169 - 144 } { 169 } } = \sqrt { \frac { 25 } { 169 } }$
$ \Rightarrow \quad \sin x = \frac { 3 } { 5 } \text { and } \sin y = \frac { 5 } { 13 }$
We know that, cos(x + y) = cosx cosy - sinx siny
$ \Rightarrow \quad \cos ( x + y ) = \left( \frac { 4 } { 5 } \times \frac { 12 } { 13 } \right) - \left( \frac { 3 } { 5 } \times \frac { 5 } { 13 } \right)$
$ \Rightarrow \quad \cos ( x + y ) = \frac { 48 } { 65 } - \frac { 15 } { 65 }$
$ \Rightarrow \quad \cos ( x + y ) = \frac { 33 } { 65 }$
$ \Rightarrow \quad x + y = \cos ^ { - 1 } \frac { 33 } { 65 }$
$ \therefore \quad \cos ^ { - 1 } \frac { 4 } { 5 } + \cos ^ { - 1 } \frac { 12 } { 13 } = \cos ^ { - 1 } \frac { 33 } { 65 }$[from Equation (i)]
Hence proved.
View full question & answer→Question 61 Mark
Prove that $\sin ^ { - 1 } \frac { 8 } { 17 } + \sin ^ { - 1 } \frac { 3 } { 5 } = \tan ^ { - 1 } \frac { 77 } { 36 }.$
AnswerWe need to prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$
Let us consider, $\sin ^{-1} \frac{8}{17}=x$ and $\sin ^{-1} \frac{3}{5}=y, \quad x, y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $\Rightarrow \quad \sin x=\frac{8}{17}$ and $\sin y=\frac{3}{5}$
Now, $\cos ^2 x=1-\sin ^2 x=1-\frac{64}{289}=\frac{225}{289}$
$\Rightarrow \quad \cos x=\sqrt{\frac{225}{289}}=\frac{15}{17}$ [taking positive square root as $\mathrm{x} \quad x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ] and $\cos ^2 y=1-\sin ^2 y=1-\frac{9}{25}=\frac{16}{25}$
$\Rightarrow \quad \cos y=\sqrt{\frac{16}{25}}=\frac{4}{5}$ [taking positive square root as $\left.x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$
Clearly, $\tan \mathrm{x}=\frac{\sin x}{\cos x}=\frac{8}{15}$ and $\tan y=\frac{3}{4}$
$\Rightarrow \quad x=\tan ^{-1} \frac{8}{15}$ and $y=\tan ^{-1} \frac{3}{4}$
Now, in general we can write LHS $=x+y=\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4}$
$=\tan ^{-1}\left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right)$
$=\tan ^{-1}\left(\frac{32+45}{60-24}\right)=\tan ^{-1}\left(\frac{77}{36}\right)=R H S$
Hence Proved.
View full question & answer→Question 71 Mark
Prove that: $2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$
AnswerLet ${\sin ^{ - 1}}\frac{3}{5} = \theta$ so that $\sin \theta = \frac{3}{5}$
$\therefore \;\cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$\therefore \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{3}{4}$
Since $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$ = \frac{{2 \times \frac{3}{4}}}{{1 - \frac{9}{{16}}}} = \frac{{\frac{3}{2}}}{{\frac{7}{{16}}}} = \frac{{24}}{7}$
$\Rightarrow 2\theta = {\tan ^{ - 1}}\frac{{24}}{7}$
$ \Rightarrow 2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$
View full question & answer→Question 81 Mark
Find the value of ${\tan ^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right)$
Answer${\tan ^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right)$ $\ne\frac{7π}{6}$ as principal value branch of $\tan^{-1}$ is $(\frac{-π}{2},\frac{π}{2})$ = So, $tan^{-1}(\frac{7π}{6})$
= ${\tan ^{ - 1}}\left[ {\tan \left( {\pi + \frac{\pi }{6}} \right)} \right]$
$$= ${\tan ^{ - 1}}\left( {\tan \frac{{\pi }}{6}} \right)$
=$\frac{\pi}{6}$ $$ $$
$\therefore$ $tan^{-1}(tan\frac{7π}{6})$ = $\frac{π}{6}$.
View full question & answer→Question 91 Mark
$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ then x is equal to
Answer$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
Now, we will put x = sin y in the given equation, and we get
$\sin ^{-1}(1-\sin y)-2 \sin ^{-1} \sin y=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(1-\sin y)-2 y=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(1-\sin y)=\frac{\pi}{2}+2 y$
$\Rightarrow 1-\sin y=\sin \left(\frac{\pi}{2}+2 y\right)$
$\Rightarrow 1-\sin y=\cos 2 y \text ({ as } \sin \left(\frac{\pi}{2}+x\right)=\cos x)$
$\Rightarrow$ $1 - cos 2y = sin y$
$\Rightarrow$ $2~sin2 y = sin y$
$\Rightarrow$ $sin y .(2 sin y - 1) = 0$
$\Rightarrow$ $sin y = 0 ~~or~~siny= \frac 12 $
$\therefore$ $x = 0~~ or~~x=\frac 12 $
Now, if we put x = $\frac{1}{2}$, then we will see that,
L.H.S. = $\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$
= $\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$
= $-\sin ^{-1} \frac{1}{2}$
= $-\frac{\pi}{6} \neq \frac{\pi}{2} \neq$ R.H.S
Hence, x = $\frac{1}{2}$ is not the solution of the given equation.
Thus, x = 0
View full question & answer→Question 101 Mark
$\sin (\tan^{–1} x), | x | < 1$ is equal to
AnswerLet $\tan^{-1} x = y,$ then $\tan y = x$
$\Rightarrow \sin y = \frac{x}{\sqrt{1+x^{2}}} $
$ \therefore y = \sin^{-1} \left(\frac{x}{\sqrt{1+x^{2}}}\right) $
$ \Rightarrow \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) $
$ \Rightarrow \sin (\tan^{-1} x) = \sin \left(\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right) $
$= \frac{x}{\sqrt{1+x^{2}}}$
View full question & answer→Question 111 Mark
Solve the equation: ${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\tan ^{ - 1}}x,(x > 0)$
AnswerPut $x = \tan \theta$
${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\tan ^{ - 1}}x$
$\Rightarrow {\tan ^{ - 1}}\left( {\frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right) = \frac{1}{2}{\tan ^{ - 1}}\tan \theta$
$\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\tan \frac{\pi }{4} - \tan \theta }}{{\tan \frac{\theta }{4} + \tan \theta }}} \right) = \frac{1}{2}\theta$
$ \Rightarrow {\tan ^{ - 1}}\tan \left( {\frac{\pi }{4} - \theta } \right) = \frac{\theta }{2}$
$\Rightarrow \frac{\pi }{4} - \theta = \frac{\theta }{2}$
$\Rightarrow \theta + \frac{\theta }{2} = \frac{\pi }{4}$
$\Rightarrow \frac{{3\theta }}{2} = \frac{\pi }{4}$
$\Rightarrow 12\theta = 2\pi$
$\Rightarrow \theta = \frac{\pi }{6}$
$\therefore x = \tan \theta = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$
View full question & answer→Question 121 Mark
Solve the equation: $2 \tan^{-1} (\cos x) = \tan^{-1} (2 cosec x)$
AnswerHere, we are required to find the value of $x,$
Now, the given equation is $2 \tan ^ { - 1 } ( \cos x ) = \tan ^ { - 1 } ( 2 cosec x ) $
$ \Rightarrow \tan ^ { - 1 } \left( \frac { 2 \cos x } { 1 - \cos ^ { 2 } x } \right) = \tan ^ { - 1 } \left( \frac { 2 } { \sin x } \right) \left[ \because 2 \tan ^ { - 1 } x = \tan ^ { - 1 } \left( \frac { 2 x } { 1 - x ^ { 2 } } \right) ; - 1 < x < 1 \right] $
$ \Rightarrow \frac { 2 \cos x } \sin ^ { 2 } {x } = \frac { 2 } { \sin x } \left[ \because 1 - \cos ^ { 2 } x = \sin ^ { 2 } x \right] $
$ \Rightarrow \sin x \cos\ x - \sin^{2x} = 0$
$ \Rightarrow \sin x\ (\cos\ x - \sin\ x) = 0$
$ \Rightarrow \sin x = 0 or \cos x = \sin\ x$
$ \Rightarrow \sin x = \sin 0$ or $\cot x = 1 = \cot \pi / 4 $
$ \therefore x = 0 \text { or } \frac { \pi } { 4 }$
But here at $x = 0$, the given equation does not exist.
Hence, $x = \frac { \pi } { 4 }$ is the only solution.
View full question & answer→Question 131 Mark
Prove that: $ \tan ^ { - 1 } \left[ \frac { \sqrt { 1 + x } - \sqrt { 1 - x } } { \sqrt { 1 + x } + \sqrt { 1 - x } } \right]$$ = \frac { \pi } { 4 } - \frac { 1 } { 2 } \cos ^ { - 1 } x , - \frac { 1 } { \sqrt { 2 } } \leq x \leq 1$. [Hint: Put x = cos 2$\theta$]
Answer$ \tan ^ { - 1 } \left[ \frac { \sqrt { 1 + x } - \sqrt { 1 - x } } { \sqrt { 1 + x } + \sqrt { 1 - x } } \right] = \frac { \pi } { 4 } - \frac { 1 } { 2 } \cos ^ { - 1 } x$
Let LHS = $ \tan ^ { - 1 } \left( \frac { \sqrt { 1 + x } - \sqrt { 1 - x } } { \sqrt { 1 + x } + \sqrt { 1 - x } } \right)$
Put x = $ \cos 2 \theta , $ $\therefore \theta = \frac { 1 } { 2 } \cos ^ { - 1 } x$
$ \therefore \quad \mathrm { LHS } = \tan ^ { - 1 } \left( \frac { \sqrt { 1 + \cos 2 \theta } - \sqrt { 1 - \cos 2 \theta } } { \sqrt { 1 + \cos 2 \theta } + \sqrt { 1 - \cos 2 \theta } } \right)$
$ = \tan ^ { - 1 } \left( \frac { \sqrt { 2 } \cos \theta - \sqrt { 2 } \sin \theta } { \sqrt { 2 } \cos \theta + \sqrt { 2 } \sin \theta } \right)$
$ \left[ \because 1 + \cos 2 \theta = 2 \cos ^ { 2 } \theta , 1 - \cos 2 \theta = 2 \sin ^ { 2 } \theta \right]$
$ = \tan ^ { - 1 } \left( \frac { 1 - \tan \theta } { 1 + \tan \theta } \right)$
[dividing numerator and denominator by $ \sqrt { 2 } \cos \theta$]
$ = \tan ^ { - 1 } \left[ \tan \left( \frac { \pi } { 4 } - \theta \right) \right]$
$ \left[ \because \frac { 1 - \tan \theta } { 1 + \tan \theta } = \tan \left( \frac { \pi } { 4 } - \theta \right) \right]$
$ = \frac { \pi } { 4 } - \theta$
$ = \frac { \pi } { 4 } - \frac { 1 } { 2 } \cos ^ { - 1 } x$ [ from equation (1) ]
= RHS
Hence Proved
View full question & answer→Question 141 Mark
Find the value of ${\cos ^{ - 1}}\left[ {\cos \left( {\frac{{13\pi }}{6}} \right)} \right]$
Answer${\cos ^{ - 1}}\cos \left( {\frac{{13\pi }}{6}} \right) = \frac{{13\pi }}{6}$, but $\frac{{13\pi }}{6} \notin [0,\;\pi ]$
which is principal branch of $\cos^{-1}$
Therefore, $\cos \left( {\frac{{13\pi }}{6}} \right) = \cos \left( {2\pi + \frac{\pi }{6}} \right)$$ = \frac{\pi }{6}$
View full question & answer→Question 151 Mark
$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ is equal to
Answer$\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$
= $\tan ^{-1} \sqrt{3}-\left(\pi-\cot ^{-1} \sqrt{3}\right)$
= $\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}-\pi$
= $\frac{\pi}{2}-\pi$
= $-\frac{\pi}{2}$
View full question & answer→Question 161 Mark
$\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to
Answer$\sin ^{-1}\left(-\frac{1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right),$ as $sin^{-1} (-x) = -sin^{-1} x$
We all know that the principle branch of$ ~~sin^{-1}x ~~$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $-\frac {\pi}{6} \in [-\frac {\pi}{2}, \frac {\pi}2]$
$\therefore$ $\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$
Now, $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$
Therefore, the required value of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=1$
View full question & answer→Question 171 Mark
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal
Answer$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$($For $\cos^{-1}(\cos x)$ type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
So here,
$\frac{7 \pi}{6} \notin[0, \pi]$
Now, $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$can be written as,
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$= \cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right]$
$= -\cos^{-1}\left(\cos \frac{\pi}{6}\right)$,
where $-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], [$since, $\cos(\pi + x) = -cosx]$
$= \pi - \cos^{-1} \left(\cos \frac{\pi}{6}\right)$ as $\cos^{-1}(-x) = \pi – \cos^{-1}$
$= \pi-\frac{\pi}{6}=\frac{5 \pi}{6}$
Hence,$ \cos^{\left(\cos \frac{7 \pi}{6}\right)=\frac{5 \pi}{6}}$
View full question & answer→Question 181 Mark
Find the value of the expression $\tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{3}{2}} \right)$.
AnswerPutting, ${\sin ^{ - 1}}\frac{3}{5} = x$ and ${\cot ^{ - 1}}\frac{3}{2} = y$ so that $\sin x = \frac{3}{5}$ and $\cot y = \frac{3}{2}$
Now, $\cos x = \sqrt {1 - {{\sin }^2}x}$ $= \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
And $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{3}{4}$
and $\tan y = \frac{1}{{\cot y}} = \frac{2}{3}$
$\therefore \tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{3}{2}} \right)$
= tan(x + y)
$ = \frac{{\tan x + \tan y}}{{1 - \tan x\tan y}} = \frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4} \times \frac{2}{3}}}$
$ = \frac{{\frac{{17}}{{12}}}}{{\frac{1}{2}}} = \frac{{17}}{6}$
View full question & answer→Question 191 Mark
Find the value of the expression ${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)$.
Answer${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)$
$ = {\tan ^{ - 1}}\left( {\tan \frac{{4\pi - \pi }}{4}} \right)$
$ = {\tan ^{ - 1}}\left[ {\tan \left( {\pi - \frac{\pi }{4}} \right)} \right]$
$ = {\tan ^{ - 1}}\left[ { - \tan \frac{\pi }{4}} \right]$
$ = {\tan ^{ - 1}}\tan \left( { - \frac{\pi }{4}} \right)$
$ = - \frac{\pi }{4}$
View full question & answer→Question 201 Mark
Find the value of the expression ${\sin ^{ - 1}}\;\sin \left( {\frac{{2\pi }}{3}} \right)$.
Answer${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) \ne \frac{{2\pi }}{3}$
as $\frac{{2\pi }}{3} \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$
${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left[ {{{\sin }^{ - 1}}\left( {\pi - \frac{\pi }{3}} \right)} \right]$
$= {\sin ^{ - 1}}\left[ {\sin \left( {\frac{\pi }{3}} \right)} \right]$
$ = \frac{\pi }{3}$
View full question & answer→Question 211 Mark
Find the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$
AnswerLet $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$,.
Then, $\cos y=-\frac{1}{\sqrt{2}}=-\cos \left(\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \left(\frac{3 \pi}{4}\right)$.
We know that the range of the principal value branch of $\cos^{-1}$ is $[0, \pi]$ and $\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}$
Therefore, the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.
View full question & answer→Question 221 Mark
Find the principal value of $\cot ^{-1}(\sqrt{3})$.
Answer$\cot^{-1}x$ represents an angle in $(0, \pi)$ whose cotangent is $x.$
Let $x=\cot ^{-1}(\sqrt{3})$
$\Rightarrow \cot x=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$
$\Rightarrow x=\frac{\pi}{6}$
$\therefore$ Principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}.$
View full question & answer→Question 231 Mark
Find the principal value of $\sec^{-1}( {\frac{2}{{\sqrt 3 }}} )$.
AnswerLet ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = \theta$
$\sec \theta = \frac{2}{{\sqrt 3 }}$
$\theta \in [0,\;\pi ] - \left\{ {\frac{\pi }{2}} \right\}$
$\implies\sec \theta = \sec \frac{\pi }{6}$
$ = \frac{\pi }{6}$
Principal value of ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = \frac{\pi }{6}$
View full question & answer→Question 241 Mark
Find the principal value of $\tan^{-1} (-1).$
AnswerThe given inverse trignometric function is $\tan ^ { - 1 } ( - 1 )$
Now, $\tan ^ { - 1 } ( - 1 ) = \tan ^ { - 1 } \left[ - \tan (\frac { \pi } { 4 } \right)]$ $\left[ \because \tan \frac { \pi } { 4 } = 1 \right]$
$= \tan ^ { - 1 } \left[ \tan \left( - \frac { \pi } { 4 } \right) \right]$ $[ \because - \tan \theta = \tan ( - \theta ) ]$
$= - \frac { \pi } { 4 } \left[ \because \tan ^ { - 1 } ( \tan \theta ) = \theta ; \forall \theta \in \left( \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right) \right]$
which is the required principal value.
View full question & answer→Question 251 Mark
Find the principal value of ${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)$.
AnswerLet ${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \theta$
$\cos \theta = \frac{{ - 1}}{2}$
$\theta \in [0,\pi ]$
$\cos \theta = \cos \left( {\pi - \frac{\pi }{3}} \right)$
$\theta = \frac{{2\pi }}{3}$
Principal value is$\frac{{2\pi }}{3}$
View full question & answer→Question 261 Mark
Find the principal value of $\tan ^{-1}(-\sqrt{3})$.
AnswerWe know that for any $x \in R, \tan^{-1}x$ represents an angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is $x.$
Therefore,
$\tan^{-1}(-\sqrt{3})$ $=\left(\text { An angle } \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { such that } \tan \theta=-\sqrt{3}\right)$
$=-\frac{\pi}{3}$
View full question & answer→Question 271 Mark
Find the principal value of $\operatorname{cosec}^{–1} (2)$
AnswerLet $\operatorname{cosec}^{–1} 2 = x$
Therefore, $\operatorname{cosec} x = 2 =\operatorname{cosec} \left(\frac{\pi}{6}\right)$
and $\operatorname{cosec} \left(\frac{\pi}{6}\right) = 2$
We know that the principle value range of $\operatorname{cosec}^{–1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Therefore, principle value of $\operatorname{cosec}^{–1}(2)$ is $\frac{\pi}{6}$
View full question & answer→Question 281 Mark
Find the principal value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
AnswerLet $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = x$
Then, $\cos x = \frac{\sqrt{3}}{2} = \cos \left(\frac{\pi}{6}\right)$
We know that the principle value range of $\cos^{-1}$ is $[0, \pi]$
and $\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$
Therefore, the principal value of $\cos^{-1} \left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}$
View full question & answer→Question 291 Mark
$\tan ^{-1} \sqrt{3}-~\sec ^{-1}(-2)$ is equal to
AnswerLet us take
$\tan ^{-1}(\sqrt{3})=x$ Then we get,
$\tan x=\sqrt{3}=\tan \frac{\pi}{3}$
$\Rightarrow ~x=\frac{\pi}3$
We know that range of the principle value branch of $\tan^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore, $x = \tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$
Let $\sec ^{-1}(-2)=y$ then we get,
$\text { sec } \mathrm{y}=-2=-\sec \frac{\pi}{3}=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \left(\frac{2 \pi}{3}\right)$
We know that range of the principle value branch of $\sec^{-1}$ is $[0, \pi]$
Therefore, $\sec ^{-1}(-2)=\frac{2 \pi}{3}$
Hence, $\tan ^{-1} \sqrt{3}-~\sec ^{-1}(-2)$= $\frac{\pi}3$$-\frac{2 \pi}{3}= -\frac {\pi}3$
View full question & answer→Question 301 Mark
If $\sin^{-1} x = y,$ then
Answer$\sin^{-1} x = y$
$\Rightarrow \sin y = x$
We know that range of the principle value branch of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Because, we know that
$-1\le \sin y\le 1$
$\Rightarrow \sin(-\frac {\pi}{2})\le \sin y \le \sin(\frac{\pi}{2})$
Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
View full question & answer→Question 311 Mark
Find the principal value of $cosec^{-1}(-\sqrt{2})$.
AnswerLet $cosec^{-1 }(-\sqrt{2})=y$.
Then$, cosec y = -\sqrt{2}=-cosec \left(\frac{\pi}{4}\right)=cosec \left(-\frac{\pi}{4}\right).$
We know that the range of the principal value branch of
$cosec^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ and $cosec \left(-\frac{\pi}{4}\right)=-\sqrt{2}.$
Therefore, the principal value of $cosec^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}.$
View full question & answer→Question 321 Mark
Find the principal value of $\sin ^ { - 1 } \left( - \frac { 1 } { 2 } \right).$
AnswerLet $y = \sin^{-1}\left(-\frac{1}{2}\right)$
$\sin y = \frac{-1}{2}$
$\sin y = \sin \left(\frac{-\pi}{6}\right)$
Range of principal value of $\sin^{-1}$ is between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$
Hence, the principal value is $\frac{-\pi}{6}$
View full question & answer→Question 331 Mark
Find the value of ${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right)$
Answer${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right) = ?$
${\sin ^{ - 1}}\left( {\sin \frac{{3\pi }}{5}} \right) = {\sin ^{ - 1}}\left[ {\sin \left( {\pi - \frac{{3\pi }}{5}} \right)} \right]$
$=\sin^{-1}(\sin \frac{2\pi}{5}$)
$[\because {\sin ^{ - 1}}(\sin \theta )] = \theta $
When $\theta \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$
$= {\frac{{2\pi }}{5}}$
View full question & answer→Question 341 Mark
Find the principal value of ${\cot ^{ - 1}}\left( { \frac{-1}{{\sqrt 3 }}} \right)$
AnswerLet ${\cot^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right) = \theta $ $\cot \theta = \frac{{ - 1}}{{\sqrt 3 }}$
We know that $\theta \in (0,\pi )$
$\cot \theta = \cot \left( {\pi - \frac{\pi }{3}} \right)$
$\theta = \frac{{2\pi }}{3}$
Therefore, principal value of ${\cot ^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right) = \frac{{2\pi }}{3}$
View full question & answer→Question 351 Mark
Find the principal value of ${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)$.
AnswerLet ${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \theta$
$\implies\sin \theta = \frac{1}{{\sqrt 2 }}$
We know that $\theta \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$
$\implies\sin \theta = \sin \frac{\pi }{4}$
$\implies\theta = \frac{\pi }{4}$
Therefore, principal value of ${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)$ is $\frac{\pi }{4}$
View full question & answer→