Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{Given that, }}{\cos ^{ - 1}}x = 2\,{\sin ^{ - 1}}x \cr & \Rightarrow \frac{\pi }{2} - {\sin ^{ - 1}}x = 2\,{\sin ^{ - 1}}x \cr & \Rightarrow \frac{\pi }{2} = 3\,{\sin ^{ - 1}}x \cr & \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{6} \cr & {\text{So}},x = \sin \frac{\pi }{6} = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Given, }}{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2} ......(1)\cr & {\text{and }}{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = 0 \cr & \Rightarrow \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right) - \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}y} \right) = 0 \cr & \Rightarrow {\sin ^{ - 1}}y - {\sin ^{ - 1}}x = 0 ......(2) \cr & \Rightarrow {\sin ^{ - 1}}y = {\sin ^{ - 1}}x \cr} $$
From equations (1) and (2) , we get
$$\eqalign{ & 2{\sin ^{ - 1}}x = \frac{\pi }{2} \cr & \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{4} \cr & \Rightarrow x = \frac{1}{{\sqrt 2 }} \cr} $$
From equation (ii)
$$y = \frac{1}{{\sqrt 2 }}$$

$$\eqalign{ & {\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a \cr & - \frac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \frac{\pi }{2}; \cr & \therefore \,\, - \frac{\pi }{2} \leqslant 2{\sin ^{ - 1}}a \leqslant \frac{\pi }{2} \cr & - \frac{\pi }{4} \leqslant {\sin ^{ - 1}}a \leqslant \frac{\pi }{4}\,\,{\text{or }}\frac{{ - 1}}{{\sqrt 2 }} \leqslant a \leqslant \frac{1}{{\sqrt 2 }} \cr & \therefore \,\,\left| a \right| \leqslant \frac{1}{{\sqrt 2 }} \cr} $$

$$2{\tan ^{ - 1}}\frac{1}{5} = {\tan ^{ - 1}}\frac{{2 \cdot \frac{1}{5}}}{{1 - {{\left( {\frac{1}{5}} \right)}^2}}} = {\tan ^{ - 1}}\frac{5}{{12}}$$
∴ the value $$ = \frac{{\tan \left( {{{\tan }^{ - 1}}\frac{5}{{12}}} \right) - \tan \frac{\pi }{4}}}{{1 + \tan \left( {{{\tan }^{ - 1}}\frac{5}{{12}}} \right) \cdot \tan \frac{\pi }{4}}} = \frac{{\frac{5}{{12}} - 1}}{{1 + \frac{5}{{12}} \cdot 1}}.$$

$$\eqalign{ & {\text{Since, }}0 \leqslant {\cos ^{ - 1}}{x_i} \leqslant \pi , \cr & \therefore {\cos ^{ - 1}}{x_i} = 0{\text{ for all }}i. \cr & \therefore {x_i} = 1{\text{ for all }}i \cr & \therefore \sum\limits_{i = 1}^{2n} {{x_i} = 2n} \cr} $$

$${\sin ^{ - 1}}x,{\cos ^{ - 1}}x\,$$   exist for $$ - 1 \leqslant x \leqslant 1.$$   But for $${\sin ^{ - 1}}\left( {1 - x} \right)$$   we must have $$ - 1 \leqslant 1 - x \leqslant 1,\,{\text{i}}{\text{.e}}{\text{., }}0 \leqslant x \leqslant 2.$$      So the equation may hold for $$0 \leqslant x \leqslant 1.$$  Therefore, the options $$A, B$$  and $$C$$ are incorrect.

$$\eqalign{ & {\text{Let, }}{\sin ^{ - 1}}x = {\tan ^{ - 1}}y = \theta \cr & \Rightarrow x = \sin \theta {\text{ and }}y = \tan \theta \cr & \frac{1}{{{x^2}}} = \frac{1}{{{{\sin }^2}\theta }} = {\text{cose}}{{\text{c}}^2}\theta \cr & {\text{and }}\frac{1}{{{y^2}}} = \frac{1}{{{{\tan }^2}\theta }} = {\cot ^2}\theta . \cr & \Rightarrow \frac{1}{{{x^2}}} - \frac{1}{{{y^2}}} = {\text{cose}}{{\text{c}}^2}\theta - {\cot ^2}\theta = 1 \cr} $$

No explanation is given for this question. Let's discuss the answer together.

Adding, $$\frac{\pi }{2} + {\sin ^{ - 1}}y - {\cos ^{ - 1}}y = \frac{\pi }{3}$$
or, $$\pi - 2{\cos ^{ - 1}}y = \frac{\pi }{3}\,\,{\text{or, }}{\cos ^{ - 1}}y = \frac{\pi }{3}$$
$$\eqalign{ & \Rightarrow \,\,y = \frac{1}{2} \cr & \therefore \,\,{\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{2} = \frac{{2\pi }}{3}\,\,{\text{or,}}\,\,{\sin ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{3} = \frac{\pi }{3} \cr & \Rightarrow \,\,x = \frac{{\sqrt 3 }}{2}. \cr} $$
So, there is one solution.

$$\eqalign{ & {\text{Let, }}{\sec ^2}\left( {{{\tan }^{ - 1}}\left( {\frac{5}{{11}}} \right)} \right) \cr & = 1 + {\tan ^2}\left( {{{\tan }^{ - 1}}\left( {\frac{5}{{11}}} \right)} \right)\left( {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) \cr & = 1 + {\left[ {\tan \left( {{{\tan }^{ - 1}}\left( {\frac{5}{{11}}} \right)} \right)} \right]^2} = 1 + {\left( {\frac{5}{{11}}} \right)^2} \cr & = 1 + \frac{{25}}{{121}} = \frac{{146}}{{121}} \cr} $$