Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

We know that $$\cot A > 1{\text{ if }}0 < A < \frac{\pi }{4}$$
$$\eqalign{ & {\text{and }}\cot A < 1{\text{ if }}\frac{\pi }{4} < A < \frac{\pi }{2} \cr & {\tan ^{ - 1}}\left( {\cot A} \right) + {\tan ^{ - 1}}\left( {{{\cot }^3}A} \right) = \pi + {\tan ^{ - 1}}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}} \cr & {\text{If }}0 < A < \frac{\pi }{4}{\text{ and }} = {\tan ^{ - 1}}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}}{\text{ if }}\frac{\pi }{4} < A < \frac{\pi }{2} \cr & {\text{Also}},{\text{ }}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}} = \frac{{\cot A\,{\text{cose}}{{\text{c}}^2}A \cdot {{\sin }^4}A}}{{{{\sin }^4}A - {{\cot }^4}A}} \cr & = \frac{{\sin A\cos A}}{{\left( {{{\sin }^2}A + {{\cos }^2}A} \right)\left( {{{\sin }^2}A - {{\cos }^2}A} \right)}} = - \frac{{\sin 2A}}{{2\cos 2A}} = - \frac{1}{2}\tan 2A \cr} $$
Hence, $${\tan ^{ - 1}}\left( {\frac{1}{2}\tan 2A} \right) + {\tan ^{ - 1}}\left( {\cot A} \right) + {\tan ^{ - 1}}\left( {{{\cot }^3}A} \right) = \pi ,$$
\[ = \left\{ {\begin{array}{*{20}{c}} {\pi \,\,\,\,\,\,\,{\rm{ if }}\, 0 < A < \frac{\pi }{4}}\\ {0\,\,\,\,\,\,\,{\rm{ if }}\,\frac{\pi }{4} < A < \frac{\pi }{2}} \end{array}\,\,\,\,\,\left[ {{\rm{Since,}}\,{{\tan }^{ - 1}}\left( { - x} \right) = - {{\tan }^{ - 1}}x} \right]} \right.\]

$$\eqalign{ & {\text{Let, }}\sqrt {\tan \alpha } = \tan x, \cr & {\text{then }}u = {\cot ^{ - 1}}\left( {\tan x} \right) - {\tan ^{ - 1}}\left( {\tan x} \right) = \frac{\pi }{2} - x - x = \frac{\pi }{2} - 2x \cr & \Rightarrow 2x = \frac{\pi }{2} - u \cr & \Rightarrow x = \frac{\pi }{4} - \frac{u}{2} \cr & \Rightarrow \tan x = \tan \left( {\frac{\pi }{4} - \frac{u}{2}} \right) \cr & \Rightarrow \sqrt {\tan \alpha } = \tan \left( {\frac{\pi }{4} - \frac{u}{2}} \right) \cr} $$

$$0 \leqslant {\cos ^{ - 1}}x \leqslant \pi .$$    Hence, from the question, $${\cos^{ - 1}}{x_i} = 0$$   for all $$i.$$
$$\therefore {x_i} = 1$$  for all $$i.$$

$$\eqalign{ & {\text{Let, }}\sin \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{5}} \right) + {{\cos }^{ - 1}}x} \right] = 1 \cr & \Rightarrow {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) + {\cos ^{ - 1}}x = {\sin ^{ - 1}}1 \cr & \Rightarrow {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) + {\cos ^{ - 1}}x = \frac{\pi }{2} \cr & \Rightarrow {\cos ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\cos ^{ - 1}}\left( {\frac{1}{5}} \right) \cr & \Rightarrow x = \frac{1}{5} \cr} $$

$$f\left( x \right) = {\sin ^{ - 1}}x + {\tan ^{ - 1}}x + {\sec ^{ - 1}}x;$$       clearly, the domain of $$f\left( x \right)$$  is $$x = \pm 1.$$   Thus the range is $$\left\{ {f\left( 1 \right),f\left( { - 1} \right)} \right\}{\text{, i}}{\text{.e}}{\text{., }}\left\{ {\frac{\pi }{4},\frac{{3\pi }}{4}} \right\}.$$

$$\eqalign{ & {\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x \cr & {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt {\cos \alpha } }}} \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x \cr & \Rightarrow \,\,{\tan ^{ - 1}}\frac{{\frac{1}{{\sqrt {\cos \alpha } }} - \sqrt {\cos \alpha } }}{{1 + \frac{1}{{\sqrt {\cos \alpha } }}.\sqrt {\cos \alpha } }} = x \cr & \Rightarrow \,\,{\tan ^{ - 1}}\frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }} = x \cr & \Rightarrow \,\,\tan x = \frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }}\,\,{\text{or }}\cot x = \frac{{2\sqrt {\cos \alpha } }}{{1 - \cos \alpha }} \cr & \Rightarrow \,\,\sin x = \frac{{1 - \cos \alpha }}{{1 + \cos \alpha }} \cr & = \frac{{1 - \left( {1 - 2{{\sin }^2}\frac{\alpha }{2}} \right)}}{{1 + 2{{\cos }^2}\frac{\alpha }{2} - 1}}\,\,{\text{or }}\sin x = {\tan ^2}\frac{\alpha }{2} \cr} $$

$$\eqalign{ & \theta = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\tan }^2}\theta - \frac{1}{3}\tan \theta }}{{1 + \frac{2}{3}{{\tan }^3}\theta }}} \right] \cr & \Rightarrow \tan \theta = \frac{{6\,{{\tan }^2}\theta - \tan \theta }}{{3 + 2\,{{\tan }^3}\theta }} \cr & \Rightarrow 1 = \frac{{6\tan \theta - 1}}{{3 + 2\,{{\tan }^3}\theta }}{\text{ or }}\tan \theta = 0 \cr & \Rightarrow 2\,{\tan ^3}\theta - 6\tan \theta + 4 = 0 \cr & \Rightarrow {\left( {\tan \theta - 1} \right)^2}\left( {\tan \theta + 2} \right) = 0 \cr & \Rightarrow \tan \theta = 1;\tan \theta = - 2;\tan \theta = 0. \cr} $$