Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{We have}},1 \leqslant {\sin ^{ - 1}}{\cos ^{ - 1}}{\sin ^{ - 1}}{\tan ^{ - 1}}x \leqslant \frac{\pi }{2} \cr & \Rightarrow \sin 1 \leqslant {\cos ^{ - 1}}{\sin ^{ - 1}}{\tan ^{ - 1}}x \leqslant 1 \cr & \Rightarrow \cos \sin 1 \geqslant {\sin ^{ - 1}}{\tan ^{ - 1}}x \geqslant \cos 1 \cr & \Rightarrow \sin \cos \sin 1 \geqslant {\tan ^{ - 1}}x \geqslant \sin \cos 1 \cr & \Rightarrow \tan \sin \cos \sin 1 \geqslant x \geqslant \tan \sin \cos 1 \cr & \therefore x \in \left[ {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right] \cr} $$

$$\eqalign{ & {\text{In }}\left[ {0,\pi } \right],\left| y \right| = \sin x,y = {\cos ^{ - 1}}\left( {\cos x} \right) = x. \cr & {\text{In }}\left[ {\pi ,2\pi } \right],\left| y \right| = \sin x,y = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - x} \right)} \right\} = 2\pi - x. \cr & {\text{In }}\left[ { - \pi ,0} \right],\left| y \right| = \sin x,y = {\cos ^{ - 1}}\left\{ {\cos \left( { - x} \right)} \right\} = - x. \cr & {\text{In }}\left[ { - 2\pi , - \pi } \right],\left| y \right| = \sin x,y = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi + x} \right)} \right\} = 2\pi + x. \cr} $$
Plotting the graphs, we have,


∴ there are 3 solutions $$\left( {0,0} \right),\left( {2\pi ,0} \right),\left( { - 2\pi ,0} \right).$$

We know that $$0 \leqslant {\cos ^{ - 1}}x \leqslant \pi .$$
Hence, from the question
$${\cos ^{ - 1}}\lambda = \pi ,{\cos ^{ - 1}}\mu = \pi ,{\cos ^{ - 1}}\gamma = \pi $$
[ $$\because {\cos ^{ - 1}}\lambda + {\cos ^{ - 1}}\mu + {\cos ^{ - 1}}\gamma = 3\pi $$       is possible only when each term attains its maximum. ]
$$\eqalign{ & \Rightarrow \lambda = \mu = \gamma = - 1 \cr & \Rightarrow \lambda \mu + \mu \gamma + \gamma \lambda = 3. \cr} $$

$$\eqalign{ & {\text{Let, }}\theta = {\text{cose}}{{\text{c}}^{ - 1}}\sqrt {\left( {{n^2} + 1} \right)\left( {{n^2} + 2n + 2} \right)} \cr & \Rightarrow {\text{cose}}{{\text{c}}^2}\theta = \left( {{n^2} + 1} \right)\left( {{n^2} + 2n + 2} \right) \cr & = {\left( {{n^2} + 1} \right)^2} + 2n\left( {{n^2} + 1} \right) + {n^2} + 1 \cr & = {\left( {{n^2} + n + 1} \right)^2} + 1 \cr & \Rightarrow {\cot ^2}\theta = {\left( {{n^2} + n + 1} \right)^2} \cr & \Rightarrow \tan \theta = \frac{1}{{{n^2} + n + 1}} = \frac{{\left( {n + 1} \right) - n}}{{1 + \left( {n + 1} \right)n}} \cr & \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\frac{{\left( {n + 1} \right) - n}}{{1 + \left( {n + 1} \right)n}}} \right] = {\tan ^{ - 1}}\left( {n + 1} \right) - {\tan ^{ - 1}}n \cr} $$
Thus, sum $$n$$ terms of the given series
$$\eqalign{ & = \left( {{{\tan }^{ - 1}}2 - {{\tan }^{ - 1}}1} \right) + \left( {{{\tan }^{ - 1}}3 - {{\tan }^{ - 1}}2} \right) + \left( {{{\tan }^{ - 1}}4 - {{\tan }^{ - 1}}3} \right) + ..... + \left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right) \cr & \Rightarrow {\tan ^{ - 1}}\left( {n + 1} \right) - \frac{\pi }{4} \cr} $$

$${\sin ^{ - 1}}\left\{ {2x\left( {1 - {x^2}} \right)} \right\} = 2\,{\sin ^{ - 1}}x$$       is true $$\forall x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$

$$\eqalign{ & \because {T_r} = {\sin ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right) \cr & {S_n} = \sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right)} \cr & = \sum\limits_{r = 1}^n {\left\{ {{{\tan }^{ - 1}}\sqrt r - {{\tan }^{ - 1}}\sqrt {\left( {r - 1} \right)} } \right\}} \cr & = {\tan ^{ - 1}}\sqrt n - {\tan ^{ - 1}}\sqrt 0 = {\tan ^{ - 1}}\sqrt n - 0 \cr & \therefore {S_\infty } = {\tan ^{ - 1}}\infty = \frac{\pi }{2} \cr} $$

Given,
$$\eqalign{ & {\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) - {\cos ^{ - 1}}\left( {\frac{{1 - {b^2}}}{{1 + {b^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) \cr & \therefore 2{\tan ^{ - 1}}a - 2{\tan ^{ - 1}}b = 2{\tan ^{ - 1}}x \cr & \Rightarrow {\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}x \cr & \Rightarrow {\tan ^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right) = {\tan ^{ - 1}}x \cr & \Rightarrow x = \frac{{a - b}}{{1 + ab}} \cr} $$

Here, $$\left| {\cos x} \right| = {\sin ^{ - 1}}\left( {\sin x} \right).$$
If, $$ - \frac{\pi }{2} \leqslant x \leqslant \frac{\pi }{2}\,{\text{then}}\,\,\cos x = x.$$
In the case there is one solution, obtained graphically.

If, $$\frac{\pi }{2} < x \leqslant \pi $$   then $$ - \cos x = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - x} \right)} \right\} = \pi - x.$$
$$\therefore \,\,\cos x = x - \pi .$$
In this case there is one solution, obtained graphically.

If, $$ - \pi \leqslant x < - \frac{\pi }{2}$$   then $$ - \cos x = {\sin ^{ - 1}}\left\{ {\sin \left( { - \pi - x} \right)} \right\} = - x - \pi ,\,{\text{i}}{\text{.e}}{\text{.,}}\cos x = x + \pi .$$
This gives no solution as can be seen from their graphs.

$$\eqalign{ & {\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6} \cr & \Rightarrow \,\,\cot \left( {{{\cot }^{ - 1}}\frac{n}{\pi }} \right) < \cot \frac{\pi }{6} \cr & \Rightarrow \,\,\frac{n}{\pi } < \sqrt 3 \cr & \Rightarrow \,\,n < \sqrt 3 \pi = 5.5\left( {{\text{nearly}}} \right). \cr} $$
So, the maximum value of $$n$$ is $$5.$$

$$\eqalign{ & {\text{Let,}}\,\,{\sin ^{ - 1}}\left( 1 \right) + {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) = {\sin ^{ - 1}}x \cr & {\text{Let,}}\,\,{\sin ^{ - 1}}\left( 1 \right) = \theta \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow \cos \theta = 0\,\,{\text{and }}{\sin ^{ - 1}}\left( {\frac{4}{5}} \right) = \phi \cr & \Rightarrow \sin \phi = \left( {\frac{4}{5}} \right) \cr & \Rightarrow \cos \phi = \sqrt {1 - \frac{{16}}{{25}}} = \sqrt {\frac{9}{{25}}} = \frac{3}{5} \cr & \therefore {\sin ^{ - 1}}x = \theta + \phi \cr & \Rightarrow x = \sin \left( {\theta + \phi } \right) = \sin \theta \cos \phi + \cos \theta \sin \phi \cr & = 1 \times \frac{3}{5} + 0 \times \frac{4}{5} \cr & \Rightarrow x = \frac{3}{5} \cr} $$