Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

Clearly, $$x\left( {x + 1} \right) \geqslant 0$$   and $${x^2} + x + 1 \leqslant 1.$$    Together they imply $$x\left( {x + 1} \right) = 0.$$
∴ $$x = 0, -1.$$   When $$x = 0,$$  L.H.S. $$ = {\tan^{ - 1}}0 + {\sin ^{ - 1}}1 = \frac{\pi }{2}.$$
When $$x = -1,$$   L.H.S. $$ = {\tan^{ - 1}}0 + {\sin ^{ - 1}}\sqrt {1 - 1 + 1} = 0 + {\sin ^{ - 1}}1 = \frac{\pi }{2}.$$

$$\eqalign{ & {\sin ^{ - 1}}\left( {\log \left[ x \right]} \right){\text{is defined if}}\, - 1 \leqslant \log \left[ x \right] \leqslant 1\,{\text{and }}\left[ x \right] > 0 \cr & \Rightarrow \frac{1}{e} \leqslant \left[ x \right] \leqslant e \cr & \Rightarrow \left[ x \right] = 1,2 \cr & \Rightarrow x \in \left[ {1,3} \right) \cr & {\text{Again}},\,\,\log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right){\text{ is defined if}} \cr & {\sin ^{ - 1}}\left[ x \right] > 0{\text{ and }} - 1 \leqslant \left[ x \right] \leqslant 1 \cr & \Rightarrow \left[ x \right] > 0{\text{ and }} - 1 \leqslant \left[ x \right] \leqslant 1 \cr & \Rightarrow 0 < \left[ x \right] \leqslant 1 \cr & \Rightarrow x \in \left[ {1,2} \right) \cr & \therefore {\text{Domain of }}f\left( x \right) = \left[ {1,2} \right) \cr & {\text{For }}1 \leqslant x < 2,\left[ x \right] = 1 \cr & \therefore f\left( x \right) = {\sin ^{ - 1}}0 + \log \frac{\pi }{2} = \log \frac{\pi }{2},\forall x \in \left[ {1,2} \right) \cr & \therefore {\text{Range of }}f\left( x \right) = \left\{ {\log \frac{\pi }{2}} \right\} \cr} $$

Clearly, $$0 \leqslant {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \leqslant \frac{\pi }{2}$$    because $${\cos ^{ - 1}}x$$  is in the first quadrant when $$x$$ is positive.
$$\therefore \,\,{\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \geqslant 0.\,{\text{So,}}\,{\cos ^{ - 1}}\frac{x}{2} \geqslant {\cos ^{ - 1}}x.$$
Also $$\left| {\frac{x}{2}} \right| \leqslant 1,\left| x \right| \leqslant 1.$$   This means $$\left| x \right| \leqslant 1.$$
In $$0 \leqslant x \leqslant 1,$$   clearly $${\cos^{ - 1}}\frac{x}{2} \geqslant {\cos ^{ - 1}}x$$    because $${\cos ^{ - 1}}\alpha $$  increases as $$\alpha $$ decreases.

$$\eqalign{ & 6{x^2} + 11x + 3 = 0 \cr & \Rightarrow \,\,x = - \frac{1}{3}, - \frac{3}{2}\,{\text{and}} - 1 < - \frac{1}{3} < 1, - \frac{3}{2} < - 1. \cr} $$
$$\therefore \,\,{\cos^{ - 1}}\left( { - \frac{1}{3}} \right)$$   exists but $${\cos^{ - 1}}\left( { - \frac{3}{2}} \right)$$   does not;
$${\text{cose}}{{\text{c}}^{ - 1}}\left( { - \frac{3}{2}} \right)$$   exists but $${\text{cose}}{{\text{c}}^{ - 1}}\left( { - \frac{1}{3}} \right)$$   does not;
$${\cot ^{ - 1}}\left( { - \frac{1}{3}} \right)$$   and $${\cot ^{ - 1}}\left( { - \frac{3}{2}} \right)$$   exist.

$$0 \leqslant {\cos ^{ - 1}}x \leqslant \pi .$$    Hence, from the question, $${\cos^{ - 1}}\lambda = \pi ,{\cos ^{ - 1}}\mu = \pi ,{\cos ^{ - 1}} \nu = \pi $$
$$\therefore \,\,\lambda = \mu = \nu = - 1.$$

The sum of the two given angles is
$$\eqalign{ & {\cot ^{ - 1}}\left( 2 \right) + {\cot ^{ - 1}}\left( 3 \right) = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right) \cr & = {\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4} \cr} $$
So, the third angle is $$\pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$$

$$\eqalign{ & {\tan ^{ - 1}}\sqrt {\left[ {x\left( {x + 1} \right)} \right]} = \frac{\pi }{2} - {\sin ^{ - 1}}\sqrt {\left( {{x^2} + x + 1} \right)} \cr & \Rightarrow \,\,{\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} = {\cos ^{ - 1}}\sqrt {{x^2} + x + 1} \cr & \Rightarrow \,\,{\cos ^{ - 1}}\frac{1}{{\sqrt {{x^2} + x + 1} }} = {\cos ^{ - 1}}\sqrt {{x^2} + x + 1} \cr & \Rightarrow \,\,{x^2} + x + 1 = 1 \cr & \Rightarrow \,\,x\left( {x + 1} \right) = 0 \cr} $$
$$ \Rightarrow \,\,x = 0, - 1$$   are the only real solutions.

We have,
$$\eqalign{ & {\sin ^{ - 1}}x + {\cos ^{ - 1}}x + {\tan ^{ - 1}}x = \frac{\pi }{2} + {\tan ^{ - 1}}x \cr & {\text{Now, }}{\sin ^{ - 1}}x\,\,{\text{and }}\,{\cos ^{ - 1}}x\,\,{\text{defined only if}} - 1 \leqslant x \leqslant 1 \cr & {\text{So, }} - \frac{\pi }{4} \leqslant {\tan ^{ - 1}}x \leqslant \frac{\pi }{4} \cr & \Rightarrow \frac{\pi }{4} \leqslant \frac{\pi }{2} + {\tan ^{ - 1}}x \leqslant \frac{{3\pi }}{4} \cr & \therefore k = \frac{\pi }{4}{\text{ and }}\,K = \frac{{3\pi }}{4} \cr} $$

$$\eqalign{ & {\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{5}} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - \frac{{3\pi }}{5}} \right)} \right\} = {\cos ^{ - 1}}\cos \left( {\frac{{3\pi }}{5}} \right) = \frac{{3\pi }}{5}. \cr & {\sin ^{ - 1}}\left( {\sin \frac{{7\pi }}{5}} \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi + \frac{{2\pi }}{5}} \right)} \right\} \cr & {\sin ^{ - 1}}\left( {\sin \frac{{7\pi }}{5}} \right) = {\sin ^{ - 1}}\left\{ { - \sin \frac{{2\pi }}{5}} \right\} = {\sin ^{ - 1}}\left\{ {\sin \left( { - \frac{{2\pi }}{5}} \right)} \right\} = - \frac{{2\pi }}{5}. \cr & {\sec ^{ - 1}}\left( {\sec \frac{{7\pi }}{5}} \right) = {\sec ^{ - 1}}\left\{ {\sec \left( {2\pi - \frac{{3\pi }}{5}} \right)} \right\} = {\sec ^{ - 1}}\left( {\sec \frac{{3\pi }}{5}} \right) = \frac{{3\pi }}{5}. \cr} $$

$$\eqalign{ & {\text{Given}}:{\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4} \cr & \Rightarrow {\tan ^{ - 1}}\frac{{\left( {2x + 3x} \right)}}{{\left( {1 - 2x \cdot 3x} \right)}} = {\tan ^{ - 1}}\left( 1 \right) \cr & \Rightarrow \frac{{5x}}{{1 - 6{x^2}}} = 1 \cr & \Rightarrow 6{x^2} + 5x - 1 = 0 \cr & \Rightarrow \left( {6x - 1} \right)\left( {x + 1} \right) = 0 \cr & \Rightarrow x = \frac{1}{6}{\text{ or }} - 1. \cr} $$