2 Marks Questions

Question 12 Marks

Find the value of ${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)} \right]$.

Answer

${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\left( {\sin \frac{\pi }{6}} \right)} \right)} \right]$=${\tan ^{ - 1}}\left[ {2\cos \left( {2\frac{\pi }{6}} \right)} \right]$
$={\tan ^{ - 1}}\left( {2\cos \frac{\pi }{3}} \right)$
$={\tan ^{ - 1}}\left( {2.\frac{1}{2}} \right)$
$=\tan^{-1}(1)$
$={\tan ^{ - 1}}\left( {\tan \frac{\pi }{4}} \right)$
$ = \frac{\pi }{4}$

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Question 22 Marks

Write the function in the simplest form: ${\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }},\left| x \right| < a$

Answer

$$ Put $x = a\sin \theta$ so that $\theta = {\sin ^{ - 1}}\frac{x}{a}$
$ \Rightarrow {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{\sqrt {{a^2}(1 - {{\sin }^2}\theta } }}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{\sqrt {{a^2}{{\cos }^2}\theta } }}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{a\cos \theta }}} \right)$
$= {\tan ^{ - 1}}\tan \theta$
$= \theta = {\sin ^{ - 1}}\frac{x}{a}$

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Question 32 Marks

Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),-\frac{\pi}{4}<x<\frac{3\pi}{4}$

Answer

We have,
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-x\right)\right\}$
$=\frac{\pi}{4}-x$

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Question 42 Marks

Write the function in the simplest form: ${\tan ^{ - 1}}\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} ,\;0<x < \pi $

Answer

${\tan ^{ - 1}}\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}}$
$= {\tan ^{ - 1}}\sqrt {\frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}}$
$ = {\tan ^{ - 1}}\tan \frac{x}{2}$
$= \frac{x}{2}$

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Question 52 Marks

Write the function in the simplest form: ${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x},x \ne 0$.

Answer

Put $x = \tan \theta$ hence $\theta = {\tan ^{ - 1}}x$
Now, $ {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$
$ = {\tan ^{ - 1}}\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}$
$ = {\tan ^{ - 1}}\frac{{\sec \theta - 1}}{{\tan \theta }}$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\cos \theta }} - 1}}{{\frac{{\sin \theta }}{{\cos \theta }}}}} \right)$
$= {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)$
$= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$

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Question 62 Marks

Prove that: $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\frac{1}{2},1} \right]$

Answer

We know that: $\cos 3\theta = 4 {\cos ^3}\theta - 3\cos \theta$
Put $\cos \theta = x$
$\Rightarrow \theta = {\cos ^{ - 1}}x$
$\therefore \cos 3\theta = 4{x^3} - 3x$
$ \Rightarrow 3\theta = {\cos ^{ - 1}}\left( {3x - 4{x^3}} \right)$
Putting $\theta = {\cos ^{ - 1}}x$,
$3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)$ Hence Proved.

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Question 72 Marks

Prove that: $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \frac{1}{2},\frac{1}{2}} \right]$

Answer

We know that: $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
Putting $\sin \theta = x$
$ \Rightarrow \theta = {\sin ^{ - 1}}x$
$\therefore \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
$\Rightarrow \sin 3\theta = 3x - 4{x^3}$
$\Rightarrow 3\theta = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$
Putting $\theta = {\sin ^{ - 1}}x$,
$3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$
Hence Proved.

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Question 82 Marks

Find the value of $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$.

Answer

Let $\cos ^{-1}\left(\frac{1}{2}\right)=x$. Then, $\cos x=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)$.
$\therefore \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
Let $\sin ^{-1}\left(\frac{1}{2}\right)=y$. Then, $\sin y=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right)$.
$\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
$\therefore \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ $=\frac{\pi}{3}+\frac{2 \pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$

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Question 92 Marks

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Answer

Let us consider $\tan^{-1}(1) = x$ then we obtain
$\tan x = 1 = \tan\frac{\pi}{4}$
We know that range of the principle value branch of $\tan^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Thus, $\tan^{-1} (1) =$ $\frac{\pi}{4}$
Let $\cos ^{-1}\left(-\frac{1}{2}\right)=y$
$\cos y = -\frac{1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$
We know that range of the principle value branch of $\cos^{-1}$ is $[0, \pi$]
Thus, $\cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$
Let $\sin^{-1}$ $\left(-\frac{1}{2}\right)=z$
$\sin z=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right)$
We know that range of the principle value branch of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Thus, $\sin^{-1}$ $\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$
Now,we have
$\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin \left(-\frac{1}{2}\right)$
= $\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}$

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Question 102 Marks

Write $\cot^{-1}$ $\left(\frac{1}{\sqrt{x^{2}-1}}\right), x>1$ in the simplest form.

Answer

Let $x = \sec \theta$,
then $\sqrt{x^{2}-1}=\sqrt{\sec ^{2} \theta-1}$ = tan $\theta$
Therefore, $\cot ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x$ , is the simplest form.

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Question 112 Marks

Express $\tan ^{-1}( \frac{\cos x}{1-\sin x}),-\frac{3 \pi}{2}<x<\frac{\pi}{2}$ in the simplest form.

Answer

According to question, we have
$\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ = $\tan ^{-1}\left[\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}\right]$
= $\tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2}$

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Question 122 Marks

Show that
$\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \cos ^{-1} x,~~ \frac{1}{\sqrt{2}} \leq x \leq 1$

Answer

Take $x = \cos \theta$, then $\cos^{–1} x = \theta$. We have
L.H.S $= \sin^{–1} (2 x \sqrt{1-x^{2}})$
$= \sin^{–1} (2 \cos \theta \sqrt{1-\cos ^{2} \theta})$
$= \sin^{–1} (2\cos \theta \sin \theta)$
$= \sin^{–1} (\sin 2\ \theta ) = 2\ \theta$
$= 2\cos^{–1} x$

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Question 132 Marks

Show that
$\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \sin ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$

Answer

Let $x = \sin \theta$. Then $\sin^{–1} x = \theta$. We have
$\sin ^{-1}(2 x \sqrt{1-x^{2}})$
L.H.S = $\sin ^{-1}(2 \sin \theta \sqrt{1-\sin ^{2} \theta})$
$= \sin^{–1} (2\sin \theta \cos \theta)$
$= \sin^{–1} (\sin 2 \theta)$
$= 2\theta$
$= 2 \sin^{–1} x$

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