Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\tan ^2}\left( {{{\sin }^{ - 1}}x} \right) > 1 \cr & \Rightarrow \frac{\pi }{4} < {\sin ^{ - 1}}x < \frac{\pi }{2}{\text{ or }} - \frac{\pi }{2} < {\sin ^{ - 1}}x < - \frac{\pi }{4} \cr & \Rightarrow x \in \left( {\frac{1}{{\sqrt 2 }},1} \right){\text{ or }}x \in \left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cr & \Rightarrow x \in \left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cup \left( {\frac{1}{{\sqrt 2 }},1} \right) \cr} $$

$$\eqalign{ & {\tan ^{ - 1}}\frac{{xy}}{{zr}} + {\tan ^{ - 1}}\frac{{yz}}{{xr}} + {\tan ^{ - 1}}\frac{{xz}}{{yr}} \cr & = {\tan ^{ - 1}}\left[ {\frac{{\frac{{xy}}{{zr}} + \frac{{yz}}{{xr}} + \frac{{xz}}{{yr}} - \frac{{xyz}}{{{r^3}}}}}{{1 - \left( {\frac{{{x^2} + {y^2} + {z^2}}}{{{r^2}}}} \right)}}} \right] \cr & = {\tan ^{ - 1}}\infty = \frac{\pi }{2} \cr} $$

Put $$x = \tan \theta .$$

$${\tan ^{ - 1}}x$$  and $${\cot ^{ - 1}}x$$  exist for all $$x \in R\,{\cos ^{ - 1}}\left( {2 - x} \right)$$    exists if $$ - 1 \leqslant 2 - x \leqslant 1{\text{ i}}{\text{.e}}{\text{., }}1 \leqslant x \leqslant 3$$
So, the given equation holds for $$1 \leqslant x \leqslant 3$$

$$\eqalign{ & {\text{Let, }}{\tan ^{ - 1}}x = \alpha {\text{ and }}{\tan ^{ - 1}}y = \beta \cr & \Rightarrow \tan \alpha = x,\tan \beta = y \cr} $$
The given system of equations is
$$a\tan \alpha + b\sec \alpha = c{\text{ and }}a\tan \beta + b\sec \beta = c$$
$$\therefore \alpha {\text{ and }}\beta $$   are the roots of $$a\tan \theta + b\sec \theta = c$$
$$\eqalign{ & \Rightarrow {\left( {b\sec \theta } \right)^2} = {\left( {c - a\tan \theta } \right)^2} \cr & \Rightarrow \left( {{a^2} - {b^2}} \right){\tan ^2}\theta - 2ac\tan \theta + {c^2} - {b^2} = 0 \cr & \Rightarrow \tan \alpha + \tan \beta = \frac{{2ac}}{{{a^2} - {b^2}}}{\text{ and }}\tan \alpha \tan \beta = \frac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}} \cr & \Rightarrow x + y = \frac{{2ac}}{{{a^2} - {b^2}}}{\text{ and }}xy = \frac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}} \cr & \Rightarrow 1 - xy = \frac{{{a^2} - {c^2}}}{{{a^2} - {b^2}}} \cr & \Rightarrow \frac{{x + y}}{{1 - xy}} = \frac{{2ac}}{{{a^2} - {c^2}}} \cr} $$

$${\sin ^{ - 1}}\left( {\sin 4} \right) = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - 4} \right)} \right\} = \pi - 4\,\,\left( {\because \,\, - \frac{\pi }{2} < \pi - 4 < \frac{\pi }{2}} \right)$$
∴ we have, $${x^2} - kx + \pi - 4 > 0$$     for all $$x \in R.$$
$$\therefore \,\,D < 0,{\text{i}}{\text{.e}}{\text{., }}{k^2} - 4\left( {\pi - 4} \right) < 0$$
or, $${k^2} + 4\left( {4 - \pi } \right) < 0,$$    which is not true for any real $$k.$$

Let $$x = \tan \theta .$$   Then $${\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = {\sin ^{ - 1}}\left( {\sin 2\theta } \right).$$
$$\eqalign{ & \therefore \,\,2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + {\sin ^{ - 1}}\left( {\sin 2\theta } \right). \cr & {\text{If }} - \frac{\pi }{2} \leqslant 2\theta \leqslant \frac{\pi }{2},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + 2\theta = 4{\tan ^{ - 1}}x \ne {\text{independent of }}x. \cr & {\text{If }} - \frac{\pi }{2} \leqslant \pi - 2\theta \leqslant \frac{\pi }{2},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - 2\theta } \right)} \right\} = 2\theta + \pi - 2\theta = \pi = {\text{independent of }}x. \cr} $$
$$\therefore \,\,\theta \notin \left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]\,{\text{but }}\theta \in \left[ {\frac{\pi }{4},\frac{{3\pi }}{4}} \right]$$      and from the principal value of $${\tan ^1}x,\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right).\,{\text{Hence, }}\theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right).$$
$$\eqalign{ & \therefore \,\,\theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right) \cr & \Rightarrow \,\,{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = \pi . \cr & {\text{Also at, }}\theta = \frac{\pi }{4},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2 \cdot \frac{\pi }{4} + {\sin ^{ - 1}}1 = \frac{\pi }{2} + \frac{\pi }{2} = \pi . \cr} $$
∴ the given function $$ = \pi = $$  constant if $$\theta \in \left[ {\frac{\pi }{4},\frac{\pi }{2}} \right),\,{\text{i}}{\text{.e}}{\text{., }}x \in \left[ {1, + \infty } \right).$$

$$\eqalign{ & {\text{Let, }}{\sin ^{ - 1}}\frac{1}{a} = \theta ;{\sin ^{ - 1}}\frac{1}{b} = \phi {\text{ then }}{\sin ^{ - 1}}\frac{1}{x} = \theta + \phi \cr & \Rightarrow \sin {\sin ^{ - 1}}\frac{1}{x} = \sin \left( {\theta + \phi } \right) \cr & \Rightarrow \frac{1}{x} = \sin \theta \cos \phi + \cos \theta \sin \phi \cr & = \frac{1}{a}\sqrt {1 - \frac{1}{{{b^2}}}} + \sqrt {1 - \frac{1}{{{a^2}}}} \cdot \frac{1}{b} = \frac{{\sqrt {{b^2} - 1} }}{{ab}} + \frac{{\sqrt {{a^2} - 1} }}{{ab}} \cr & \Rightarrow x = \frac{{ab}}{{\sqrt {{a^2} - 1} + \sqrt {{b^2} - 1} }} \cr} $$

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

Given equation is an identity, except the range should be same for both sides
The range of $${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right){\text{ is }}\left[ { - \frac{\pi }{2},\,\frac{\pi }{2}} \right]$$
And the range of $$2{\cos ^{ - 1}}x{\text{ is }}\left[ {0,\,2\pi } \right]$$
So the common range is $$\left[ {0,\,\frac{\pi }{2}} \right]$$
$$\eqalign{ & \Rightarrow 0 \leqslant 2{\cos ^{ - 1}}x \leqslant \frac{\pi }{2} \cr & \Rightarrow 0 \leqslant {\cos ^{ - 1}}x \leqslant \frac{\pi }{4} \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \leqslant x \leqslant 1, \cr} $$
Since $${\cos^{ - 1}}$$ is decreasing function so, inequality will get reversed.