Chapter 2: Inverse Trigonometric Functions

Ans: Let’s Consider, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

As we know that ${\tan ^{ - 1}}(\tan x) = x$

If $x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, which is the principal value branch of ${\tan ^{ - 1}}x$. Here, $\dfrac{{3\pi }}{4} \notin \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

Now, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$ can be written as:

${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ { - \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)} \right] = {\tan ^{ - 1}}\left[ { - \tan \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]$

${\tan ^{ - 1}}\left( { - \tan \dfrac{\pi }{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right]$ where $ - \dfrac{\pi }{4} \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

$\therefore {\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right] = \dfrac{{ - \pi }}{4}$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{3}{5} = x$.

Then, $\sin x = \dfrac{3}{5}$

$ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}  = \dfrac{4}{5}$

$ \Rightarrow \sec x = \dfrac{5}{4}$

$\therefore \tan x = \sqrt {{{\sec }^2}x - 1}  = \sqrt {\dfrac{{25}}{{16}} - 1}  = \dfrac{3}{4}$

$\therefore x = {\tan ^{ - 1}}\dfrac{3}{4}$

$\therefore {\sin ^{ - 1}}\dfrac{3}{5} = {\tan ^{ - 1}}\dfrac{3}{4}\quad  \ldots (i)$

Therefore, $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{2}{3}} \right)\quad $ [Using (i) and (ii)]

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}}} \right)} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{9 + 8}}{{12 - 6}}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{17}}{6}} \right) = \dfrac{{17}}{6}$

Ans: We know that ${\cos ^{ - 1}}(\cos x) = x$ if $x \in [0,\pi ]$, which is the principal value branch of ${\cos ^{ - 1}}x$. Here, $\dfrac{{7\pi }}{6} \notin [0,\pi ]$.

Now, ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ can be written as:

${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{ - 7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi  - \dfrac{{7\pi }}{6}} \right)} \right]\quad [\because $

$ + x) = \cos x]$

$\therefore {\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{6}} \right) = \dfrac{{5\pi }}{6}$

The correct answer is ${\text{B}}$.

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = x$.

Then, $\sin x = \dfrac{{ - 1}}{2} =  - \sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$. ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{\pi }{6}$

$\therefore \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{{3\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$

The correct answer is ${\text{D}}$.

Ans:

Suppose that,

$A=\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$

As we know that,

$\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$

So,

$\cot ^{-1}(-\sqrt{3})=\pi-\cot ^{-1}(\sqrt{3})$

Put the value in the equation (i),

$A=\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})$

As we know that,

$\cot ^{-1} x=\dfrac{\pi}{2}-\tan ^{-1} x$

So,

$\cot ^{-1}(\sqrt{3})=\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

Put the value in the equation (ii),

$A=\tan ^{-1}(\sqrt{3})-\pi+\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3}) $

$A=-\dfrac{\pi}{2}$

Hence, the correct option is B.