Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

Value $$ = {\sec ^2}\left( {\sec {\,^{ - 1}}2} \right) - 1 + {\text{cose}}{{\text{c}}^2}\left( {{\text{cosec}}{\,^{ - 1}}3} \right) - 1 = {2^2} - 1 + {3^2} - 1 = 11.$$

$$\because \frac{\pi }{2} < 4 < \frac{{3\pi }}{2},{\text{so }}{\sin ^{ - 1}}\sin 4 = {\sin ^{ - 1}}\sin \left( {\pi - 4} \right) = \pi - 4$$
The inequality becomes $${x^2} - kx + \pi - 4 > 0$$
The discriminant $$D = {k^2} - 4\left( {\pi - 4} \right) > 0$$     for all $$k,$$ that is $${x^2} - kx + \left( {\pi - 4} \right) > 0$$     can not hold for all $$x.$$

$$\eqalign{ & {\tan ^{ - 1}}\left( {\tan x} \right) = x\,\,{\text{if }} - \frac{\pi }{2} < x < \frac{\pi }{2}. \cr & \tan \frac{{15\pi }}{4} = \tan \left( {4\pi - \frac{\pi }{4}} \right) = \tan \left( { - \frac{\pi }{4}} \right). \cr} $$
∴ the value $$ = \cos \left\{ {{{\tan }^{ - 1}}\left( {\tan \frac{{ - \pi }}{4}} \right)} \right\} = \cos \left( { - \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}.$$

$$\eqalign{ & {\cos ^{ - 1}}\left( { - \sin \frac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {\frac{\pi }{2} + \frac{{7\pi }}{6}} \right)} \right\} \cr & {\cos ^{ - 1}}\left( { - \sin \frac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - \frac{{5\pi }}{3}} \right)} \right\} \cr & {\cos ^{ - 1}}\left( { - \sin \frac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \frac{\pi }{3}} \right) = \frac{\pi }{3}. \cr} $$
Remember, $${\cos^{ - 1}}\left( {\cos x} \right) = x\,\,{\text{if }}0 \leqslant x \leqslant \pi .$$

$$\eqalign{ & {\text{We have, }}{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \pi - {\sin ^{ - 1}}z \cr & {\text{or, }}x\sqrt {\left( {1 - {y^2}} \right)} + y\sqrt {\left( {1 - {x^2}} \right)} = z \cr & {\text{or, }}{x^2}\left( {1 - {y^2}} \right) = {z^2} + {y^2}\left( {1 - {x^2}} \right) - 2yz\sqrt {\left( {1 - {x^2}} \right)} \cr & {\text{or, }}{\left( {{x^2} - {z^2} - {y^2}} \right)^2} = 4{y^2}{z^2}\left( {1 - {x^2}} \right) \cr & {\text{or, }}{x^4} + {y^2} + {z^4} - 2{x^2}{z^2} + 2{y^2}{z^2} - 2{x^2}{y^2} + 4{x^2}{y^2}{z^2} - 4{y^2}{z^2} = 0 \cr & {\text{or, }}{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = 2\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right) \cr & \therefore k = 2 \cr} $$

$$\eqalign{ & {\text{We have, }}2{\left( {{{\sin }^{ - 1}}y} \right)^2} = \frac{{4n + 1}}{{16}}{\pi ^2} \cr & \Rightarrow 0 \leqslant \frac{{4n + 1}}{{32}}{\pi ^2} \leqslant \frac{{{\pi ^2}}}{4} \cr & {\text{Also, }}2\left( {{{\cos }^{ - 1}}x} \right) = \frac{{4n - 1}}{{16}}{\pi ^2} \cr & \Rightarrow - \frac{1}{4} \leqslant n \leqslant \frac{7}{4} \cr & {\text{Also, }}2\left( {{{\cos }^{ - 1}}x} \right) = \frac{{4n - 1}}{{16}}{\pi ^2} \cr & \Rightarrow 0 \leqslant \frac{{4n - 1}}{{32}}{\pi ^2} \leqslant \pi \cr & \Rightarrow \frac{1}{4} \leqslant n \leqslant \frac{8}{\pi } + \frac{1}{4} \cr & \Rightarrow n = 1 \cr} $$

$${\sec ^{ - 1}}x$$  is real when $$x \leqslant - 1$$  and $$x \geqslant 1.{\tan^{ - 1}}x$$   exists for all $$x \in R.$$

$$\eqalign{ & {\text{Let, }}\alpha = {\tan ^{ - 1}}x \cr & \Rightarrow \tan \alpha = x \cr & {\text{then }}\,\cos \alpha = \frac{1}{{\sqrt {1 + {{\tan }^2}\alpha } }} = \frac{1}{{\sqrt {1 + {x^2}} }} \cr & \Rightarrow \cos \left. {\left( {{{\tan }^{ - 1}}x} \right)} \right\} = \left\{ {\frac{1}{{\sqrt {1 + {x^2}} }}} \right\} \cr & {\text{So}},\,\,{\cot ^{ - 1}}\cos \left( {{{\tan }^{ - 1}}x} \right) = {\cot ^{ - 1}}\left\{ {\frac{1}{{\sqrt {1 + {x^2}} }}} \right\} \cr & {\text{Let, }}{\cot ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) = \beta \cr & \Rightarrow \cot \beta = \frac{1}{{\sqrt {1 + {x^2}} }} \cr & {\text{and }}\,\sin \beta = \frac{1}{{\sqrt {1 + {{\cot }^2}\beta } }} \cr & = \frac{{\sqrt {1 + {x^2}} }}{{\sqrt {{x^2} + 1 + 1} }} = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} \cr & \Rightarrow \sin \left[ {{{\cot }^{ - 1}}\left\{ {\cos \left( {{{\tan }^{ - 1}}} \right)} \right\}} \right] = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} \cr} $$

$$\eqalign{ & {x^2} - x - 2 > 0 \cr & \Rightarrow \,\,\left( {x - 2} \right)\left( {x + 1} \right) > 0 \cr} $$
$$ \Rightarrow \,\,x < - 1\,\,{\text{or, }}x > 2,$$     using sign scheme.
We know that $${\sin^{ - 1}}x$$  is defined for $$\left| x \right| \leqslant 1,{\cos ^{ - 1}}x$$    is defined for $$\left| x \right| \leqslant 1$$  and $${\sec^{ - 1}}x$$  is defined for $$\left| x \right| \geqslant 1,\,{\text{i}}{\text{.e}}{\text{., }}x \leqslant - 1\,\,{\text{or, }}x \geqslant 1.$$

Let, $${\sin ^{ - 1}}x = \theta $$
$$ \Rightarrow x = \sin \theta $$
$$\eqalign{ & {\sin ^{ - 1}}\left( {3\sin \theta - 4\,{{\sin }^3}\theta } \right) = {\sin ^{ - 1}}\sin 3\theta \cr & = 3\theta = 3\,{\sin ^{ - 1}}x \cr} $$
Equation $${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = 3\,{\sin ^{ - 1}}x$$      is true for all values of $$x$$ lying in the interval $$\left[ { - 1,1} \right].$$