Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & \sqrt {1 + \sin x} = \sin \frac{x}{2} + \cos \frac{x}{2} \cr & \sqrt {1 - \sin x} = \sin \frac{x}{2} - \cos \frac{x}{2} \cr & \left[ {{\text{for}}\,\frac{\pi }{4} \leqslant \frac{x}{2} \leqslant \frac{\pi }{2}\sin \frac{x}{2} \geqslant \cos \frac{x}{2}} \right] \cr & \therefore {\text{the expression is}} \cr & {\cot ^{ - 1}}\left( {\frac{{\sin \frac{x}{2} + \cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2}}}{{\sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}}}} \right) \cr & = {\cot ^{ - 1}}\left( {\tan \frac{x}{2}} \right) = {\cot ^{ - 1}}\cot \left( {\frac{\pi }{2} - \frac{x }{2}} \right) = \frac{{\pi - x}}{2} \cr} $$

$$\eqalign{ & {\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{y}{2} = \alpha \cr & {\cos ^{ - 1}}\left( {\frac{{xy}}{2} + \sqrt {\left( {1 - {x^2}} \right)\left( {1 - \frac{{{y^2}}}{4}} \right)} } \right) = \alpha \cr & {\cos ^{ - 1}}\left( {\frac{{xy + \sqrt {4 - {y^2} - 4{x^2} + {x^2}{y^2}} }}{2}} \right) = \alpha \cr & \Rightarrow \,\,4 - {y^2} - 4{x^2} + {x^2}{y^2} = 4{\cos ^2}\alpha + {x^2}{y^2} - 4xy\cos \alpha \cr & \Rightarrow \,\,4{x^2} + {y^2} - 4xy\cos \alpha = 4{\sin ^2}\alpha . \cr} $$

$$\eqalign{ & y = \left( {{{\cot }^{ - 1}}x} \right)\left( {{{\cot }^{ - 1}}\left( { - x} \right)} \right) \cr & = {\cot ^{ - 1}}\left( x \right)\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) \cr & {\text{Now, }}{\cot ^{ - 1}}\left( x \right){\text{ and }}\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) > 0 \cr & {\text{Using A}}{\text{.M}}{\text{.}} \geqslant {\text{G}}{\text{.M}}{\text{., we get}} \cr & \frac{{{{\cot }^{ - 1}}x + \left( {\pi - {{\cot }^{ - 1}}x} \right)}}{2} \geqslant \sqrt {\left( {{{\cot }^{ - 1}}x} \right)\left( {\pi - {{\cot }^{ - 1}}x} \right)} \cr & \Rightarrow 0 < {\cot ^{ - 1}}\left( x \right)\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) \leqslant \left( {\frac{{{{\cot }^{ - 1}}x + \left( {\pi - {{\cot }^{ - 1}}x} \right)}}{2}} \right) = \frac{{{\pi ^2}}}{4} \cr & \Rightarrow 0 < y \leqslant \frac{{{\pi ^2}}}{4} \cr} $$

For $${\cos ^{ - 1}}x,{\sin ^{ - 1}}x$$   to be real, $$ - 1 \leqslant x \leqslant 1.$$
But $${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$$
$$\eqalign{ & \Rightarrow \,\,2{\cos ^{ - 1}}x > \frac{\pi }{2} \cr & \Rightarrow \,\,{\cos ^{ - 1}}x > \frac{\pi }{4} \cr & \therefore \,\,\cos \left( {{{\cos }^{ - 1}}x} \right) < \cos \frac{\pi }{4}\,\,{\text{or, }}x < \frac{1}{{\sqrt 2 }}. \cr} $$
The common values are $$ - 1 \leqslant x < \frac{1}{{\sqrt 2 }}.$$

As given : $$\cos\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{2}$$
$$\eqalign{ & \Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) \cr & \Rightarrow x = \sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2} \cr & \therefore \tan\left( {{{\cos }^{ - 1}}x} \right) = \tan \left( {{{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right) \cr & = \tan \left( { \pm \frac{\pi }{6}} \right) = \pm \frac{1}{{\sqrt 3 }} \cr} $$
Hence, $$\tan\left( {{{\cos }^{ - 1}}x} \right)$$   have two values.

$$\eqalign{ & {\bf{Case 1 :}} \cr & {\text{If }}0 \leqslant x \leqslant \frac{1}{2},\,\,{\text{then }}{\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) \cr & {\cos ^{ - 1}}\left( {x \times \frac{1}{2} + \sqrt {1 - {x^2}} \frac{{\sqrt 3 }}{2}} \right) \cr & = {\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{1}{2} \cr} $$
Therefore, the equation is
$$\eqalign{ & {\cos ^{ - 1}}x + {\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{1}{2} = \frac{\pi }{3} \cr & \Rightarrow x = \frac{1}{2}. \cr & {\bf{Case 2 :}} \cr & {\text{If }}\frac{1}{2} \leqslant x \leqslant 1,{\text{ then}} \cr & {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) = {\cos ^{ - 1}}\frac{1}{2} - {\cos ^{ - 1}}x \cr} $$
Therefore, the equation is
$${\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{2} - {\cos ^{ - 1}}x = \frac{\pi }{3},$$
which is an identity.
Hence, the identity holds good for $$x \in \left[ {\frac{1}{2},1} \right].$$

$$\eqalign{ & {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2} \cr & \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}y \cr & \Rightarrow {\sin ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {y^2}} \cr & \Rightarrow {x^2} + {y^2} = 1 \cr & \Rightarrow \frac{{1 + {x^4} + {y^4}}}{{{x^2} - {x^2}{y^2} + {y^2}}} = \frac{{1 + {{\left( {{x^2} + {y^2}} \right)}^2} - 2{x^2}{y^2}}}{{1 - {x^2}{y^2}}} \cr & = \frac{{1 + 1 - 2{x^2}{y^2}}}{{1 - {x^2}{y^2}}} = 2. \cr} $$

$$\eqalign{ & \frac{1}{{1 + r + {r^2}}} = \frac{1}{{1 + r\left( {r + 1} \right)}} = \frac{{\frac{1}{{r\left( {r + 1} \right)}}}}{{1 + \frac{1}{{r\left( {r + 1} \right)}}}} = \frac{{\frac{1}{r} - \frac{1}{{r + 1}}}}{{1 + \frac{1}{r}\left( {\frac{1}{{r + 1}}} \right)}} \cr & \therefore {\tan ^{ - 1}}\left( {\frac{1}{{1 + r + {r^2}}}} \right) = {\tan ^{ - 1}}\frac{1}{r} - {\tan ^{ - 1}}\frac{1}{{r + 1}} \cr & \therefore \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}} \left( {\frac{1}{{1 + r + {r^2}}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4} \cr} $$

$${\text{cose}}{{\text{c}}^{ - 1}}\left( {\cos x} \right)$$   exists if $$\cos x \leqslant - 1$$   or $$\cos x \geqslant 1.$$   But $$ - 1 \leqslant \cos x \leqslant 1.$$
$$\therefore {\text{cose}}{{\text{c}}^{ - 1}}\left( {\cos x} \right)$$    exists if $$\cos x = 1\,{\text{or,}} - 1$$
$$ \Rightarrow \,\,x = n\pi ,n \in {\Bbb Z}.$$

$$\eqalign{ & {\text{We have,}} \cr & {\sin ^{ - 1}}\left\{ {\cot \left( {{{\sin }^{ - 1}}\sqrt {\left( {\frac{{2 - \sqrt 3 }}{4}} \right)} + {{\cos }^{ - 1}}\frac{{\sqrt {12} }}{4} + {{\sec }^{ - 1}}\sqrt 2 } \right)} \right\} \cr & = {\sin ^{ - 1}}\left\{ {\cot \left( {{{\sin }^{ - 1}}\sqrt {{{\left( {\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)}^2}} + {{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2} + {{\cos }^{ - 1}}\frac{1}{{\sqrt 2 }}} \right)} \right\} \cr & = {\sin ^{ - 1}}\left\{ {\cot \left( {{{15}^ \circ } + {{30}^ \circ } + {{45}^ \circ }} \right)} \right\} = {\sin ^{ - 1}}\left( {\cot {{90}^ \circ }} \right) \cr & = {\sin ^{ - 1}}0 = 0 \cr} $$