Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

Let, $${\cos ^{ - 1}}x = \theta .$$   Then the expression $$ = \frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}} + \frac{{1 - \tan \frac{\theta }{2}}}{{1 + \tan \frac{\theta }{2}}} = \frac{{2\left( {1 + {{\tan }^2}\frac{\theta }{2}} \right)}}{{1 - {{\tan }^2}\frac{\theta }{2}}} = \frac{2}{{\cos \theta }} = \frac{2}{x}.$$

$$\eqalign{ & {\text{Let, }}{\cos ^{ - 1}}x = y \cr & \Rightarrow x = \cos y,{\text{so that }}\frac{1}{2} \leqslant x \leqslant 1{\text{ or }}0 \leqslant y \leqslant \frac{\pi }{3} \cr & {\text{and }}\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} = \frac{1}{2}\cos y + \frac{{\sqrt 3 }}{2}\sin y \cr & = \cos \frac{\pi }{3}\cos y + \sin \frac{\pi }{3}\sin y = \cos \left( {\frac{\pi }{3} - y} \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) = \frac{\pi }{3} - y \cr} $$
$$\therefore $$ the given expression is equal to
$$y + \frac{\pi }{3} - y,{\text{ i}}{\text{.e}}{\text{., }}\frac{\pi }{3}$$

$$\eqalign{ & \because a - b < 0,{\text{ so}} \cr & {\cot ^{ - 1}}\frac{{ab + 1}}{{a - b}} = {\cot ^{ - 1}}b - {\cot ^{ - 1}}a + \pi \cr & b - c < 0,{\text{ so }}{\cot ^{ - 1}}\frac{{bc + 1}}{{b - c}} = {\cot ^{ - 1}}c - {\cot ^{ - 1}}b + \pi \cr & c - a > 0,{\text{ so }}{\cot ^{ - 1}}\frac{{ca + 1}}{{c - a}} = {\cot ^{ - 1}}a - {\cot ^{ - 1}}c \cr & {\text{Adding we get,}} \cr & {\cot ^{ - 1}}\frac{{ab + 1}}{{a - b}} + {\cot ^{ - 1}}\frac{{bc + 1}}{{b - c}} + {\cot ^{ - 1}}\frac{{ca + 1}}{{c - a}} = 2\pi \cr} $$

$$\eqalign{ & {\sin ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right) = {\tan ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right) \cr & = {\tan ^{ - 1}}\sqrt r - {\tan ^{ - 1}}\left( {\sqrt {r - 1} } \right) \cr & \Rightarrow \sum\limits_{r = 1}^n {{{\sin }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right)} \cr & = \sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}\sqrt r - {{\tan }^{ - 1}}\sqrt {r - 1} } \right)} = {\tan ^{ - 1}}\sqrt n \cr} $$

We have,
$$\eqalign{ & {\tan ^{ - 1}}\frac{x}{\pi } < \frac{\pi }{3} \cr & \Rightarrow \tan \left( {{{\tan }^{ - 1}}\frac{x}{\pi }} \right) < \tan \frac{\pi }{3} \cr & \Rightarrow \frac{x}{\pi } < \sqrt 3 \cr & \Rightarrow x < \sqrt 3 \,\pi = 5.5\left( {{\text{approx}}} \right) \cr} $$
$$\therefore $$ the maximum value of $$x$$ is $$5.$$

$$\eqalign{ & {\text{Let, }}{\cos ^{ - 1}}x = \theta \cr & \Rightarrow x = \cos \theta {\text{ or }}\sec \theta = \frac{1}{x} \cr & \Rightarrow \tan \theta = \sqrt {{{\sec }^2}\theta - 1} = \sqrt {\frac{1}{{{x^2}}} - 1} = \frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} \cr & {\text{Now, }}\sin {\cot ^{ - 1}}\tan \theta = \sin {\cot ^{ - 1}}\left( {\frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} } \right). \cr} $$
Again, putting $$x = \sin \theta ,$$   we get
$$\eqalign{ & \sin {\cot ^{ - 1}}\left( {\frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} } \right) = \sin {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \right) \cr & = \sin {\cot ^{ - 1}}\left| {\cot \theta } \right| = \sin \theta = x. \cr} $$

$$\eqalign{ & \mathop {\lim}\limits_{x \to \infty } x\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \,x\,{\tan ^{ - 1}}\left( {\frac{{\frac{{x + 1}}{{x + 2}} - \frac{x}{{x + 2}}}}{{1 + \frac{{x + 1}}{{x + 2}} \cdot \frac{x}{{x + 2}}}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } x\,{\tan ^{ - 1}}\left( {\frac{{x + 2}}{{2{x^2} + 5x + 4}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } x\left( {\frac{{{{\tan }^{ - 1}}\left( {\frac{{x + 2}}{{2{x^2} + 5x + 4}}} \right)}}{{\frac{{x + 2}}{{2{x^2} + 5x + 4}}}}} \right) \times \frac{{x\left( {x + 2} \right)}}{{2{x^2} + 5x + 4}} \cr & = 1 \times \frac{1}{2} = \frac{1}{2} \cr} $$

Value $$ = {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = {\tan ^{ - 1}}\frac{{\frac{1}{3} + \frac{1}{2}}}{{1 - \frac{1}{3} \cdot \frac{1}{2}}} = {\tan ^{ - 1}}1 = \frac{\pi }{4}.$$

Principal value branch of $${\text{cose}}{{\text{c}}^{ - 1}}x$$
$$ = \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}$$

$${\sin \frac{{5\pi }}{6}}$$  is to be written as $$\sin \alpha ,$$  where $$ - \frac{\pi }{2} \leqslant \alpha \leqslant \frac{\pi }{2}.$$
$$\therefore \,\,{\sin^{ - 1}}\left( {\sin \frac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - \frac{\pi }{6}} \right)} \right\} = {\sin ^{ - 1}}\left( {\sin \frac{\pi }{6}} \right) = \frac{\pi }{6}.$$