Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\tan ^{ - 1}}\left( {1 + x} \right) + {\tan ^{ - 1}}\left( {1 - x} \right) = \frac{\pi }{2} \cr & {\tan ^{ - 1}}\left[ {\frac{{\left( {1 + x} \right) + \left( {1 - x} \right)}}{{1 - \left( {1 + x} \right)\left( {1 - x} \right)}}} \right] = \frac{\pi }{2} \cr & \Rightarrow \frac{{1 + x + 1 - x}}{{1 - \left( {1 + x} \right)\left( {1 - x} \right)}} = \tan \frac{\pi }{2} \cr & \Rightarrow \frac{2}{{1 - \left( {1 + x} \right)\left( {1 - x} \right)}} = \frac{1}{0} \cr & \Rightarrow 1 - \left( {1 + x} \right)\left( {1 - x} \right) = 0 \cr & \Rightarrow \left( {1 + x} \right)\left( {1 - x} \right) = 1 \cr & 1 - {x^2} = 1 \cr & {x^2} = 0 \cr & x = 0 \cr} $$

$$\eqalign{ & {\text{Given, }}\,{\sin ^{ - 1}}x > {\cos ^{ - 1}}x{\text{ where}}\,x \in \left( {0,1} \right) \cr & \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{2} - {\sin ^{ - 1}}x \cr & \Rightarrow 2\,{\sin ^{ - 1}}x > \frac{\pi }{2} \cr & \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{4} \cr & \Rightarrow x > \sin \frac{\pi }{4} \cr & \Rightarrow x > \frac{1}{{\sqrt 2 }} \cr} $$
Maximum value of $${\sin ^{ - 1}}x\,\,{\text{is }}\,\frac{\pi }{2}$$
So, maximum value of $$x$$ is $$1.$$ So, $$x \in \left( {\frac{1}{{\sqrt 2 }},1} \right).$$

Put $$x = \tan \theta {\text{ and }}y = \tan \phi $$

Given that,
$$\eqalign{ & {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) + {\cos ^{ - 1}}\left( z \right) = \pi \cr & \Rightarrow {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) + {\cos ^{ - 1}}\left( z \right) = {\cos ^{ - 1}}\left( { - 1} \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) = \pi - {\cos ^{ - 1}}\left( z \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right) = {\cos ^{ - 1}}\left( { - z} \right) \cr & \Rightarrow xy - \sqrt {\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)} = - z \cr & \Rightarrow \left( {xy + z} \right) = \sqrt {\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)} \cr} $$
Squaring both sides, we get
$${x^2} + {y^2} + {z^2} + 2xyz = 1.$$

$$\eqalign{ & \tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right] \cr & = \tan \left[ {{{\tan }^{ - 1}}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}} \right] \cr & = \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4} \times \frac{2}{3}}}} \right)} \right] \cr & = \frac{{17}}{{12}} \times \frac{{12}}{6} \cr & = \frac{{17}}{6} \cr} $$

Let $$x = \sin \theta .$$   Then $$f\left( x \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\theta - \frac{\pi }{6}} \right)} \right\}.$$
$$\eqalign{ & - \frac{1}{2} \leqslant x \leqslant 1 \cr & \Rightarrow \,\, - \frac{1}{2} \leqslant \sin \theta \leqslant 1 \cr & \Rightarrow \,\, - \frac{\pi }{6} \leqslant \theta \leqslant \frac{\pi }{2}. \cr} $$
So, $$\theta - \frac{\pi }{6}$$  is in the fourth or the first quadrant. Hence, $$f\left( x \right) = \theta - \frac{\pi }{6}.$$

Since, $$x, y, z$$   are in A.P.
⇒ $$2y = x + z$$
Also, we have, $$2\,{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( z \right)$$
$$\eqalign{ & \Rightarrow {\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{x + z}}{{1 - xz}}} \right) \cr & \Rightarrow \,\,\frac{{x + z}}{{1 - {y^2}}} = \frac{{x + z}}{{1 - xz}}\,\,\left( {\because \,\,2y = x + z} \right) \cr & \Rightarrow \,\,{y^2} = xz\,\,\,{\text{or }}\,x + z = 0 \cr & \Rightarrow \,\,x = y = z \cr} $$

Let $$x = \sin \theta .$$   Then $${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta = 2{\sin ^{ - 1}}x$$          if $$ - \frac{\pi }{2} \leqslant 2\theta \leqslant \frac{\pi }{2},\,{\text{i}}{\text{.e}}{\text{.,}}\sin \left( { - \frac{\pi }{4}} \right) \leqslant \sin \theta \leqslant \sin \left( {\frac{\pi }{4}} \right),\,{\text{i}}{\text{.e}}{\text{.,}} - \frac{1}{{\sqrt 2 }} \leqslant x \leqslant \frac{1}{{\sqrt 2 }}.$$

$$\eqalign{ & {\text{We have, }}A = {\tan ^{ - 1}}2 \cr & \Rightarrow \tan A = 2\,{\text{and}}\,B = {\tan ^{ - 1}}3 \cr & \Rightarrow \tan B = 3. \cr & {\text{Since, }}A,B,C{\text{ are angles of a triangle}} \cr & \therefore A + B + C = \pi \cr & \Rightarrow C = \pi - \left( {A + B} \right)\,\,\,.....\left( 1 \right) \cr & {\text{Now}},A + B = {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3 \cr & = \pi + {\tan ^{ - 1}}\left( {\frac{{2 + 3}}{{1 - 2 \cdot 3}}} \right)\,\left[ {\because {{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = \pi + {{\tan }^{ - 1}}\frac{{x + y}}{{1 - xy}}} \right] \cr & = \pi + {\tan ^{ - 1}}\left( { - 1} \right) = \pi - {\tan ^{ - 1}}\left( { - 1} \right) \cr & = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4} \cr & \therefore {\text{from}}\,\left( 1 \right),C = \pi - \frac{{3\pi }}{4} = \frac{\pi }{4}. \cr} $$

$$\eqalign{ & {\sin ^{ - 1}}x = 2\,{\tan ^{ - 1}}x \cr & \Rightarrow {\sin ^{ - 1}}x = {\sin ^{ - 1}}\left\{ {\frac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right\} \cr & \Rightarrow \frac{{2x}}{{\left( {1 + {x^2}} \right)}} = x \cr & \Rightarrow x\left( {x + 1} \right)\left( {x - 1} \right) = 0 \cr & \Rightarrow x = \left\{ { - 1,1,0} \right\} \cr} $$