Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{we have,}}\,\,{\cot ^{ - 1}}7 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 \cr & {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{8} + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{7} + \frac{1}{8}}}{{1 - \frac{1}{7} \times \frac{1}{8}}}} \right) + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\frac{{15}}{{55}}{\tan ^{ - 1}}\frac{1}{{18}}\,\,\left( {\because \frac{1}{7} \cdot \frac{1}{8} < 1} \right) \cr & {\tan ^{ - 1}}\frac{3}{{11}} + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{{11}} + \frac{1}{{18}}}}{{1 - \frac{3}{{11}} \times \frac{1}{{18}}}}} \right)\left( {\because \frac{3}{{11}} \cdot \frac{1}{{18}} < 1} \right) \cr & = {\tan ^{ - 1}}\frac{{65}}{{195}} = {\tan ^{ - 1}}\frac{1}{3} = {\cot ^{ - 1}}3 \cr} $$

$$\eqalign{ & \left( {\frac{\pi }{2} - {{\cos }^{ - 1}}x} \right) - {\cos ^{ - 1}}x = \frac{\pi }{6}\,\,{\text{or, }}2{\cos ^{ - 1}}x = \frac{\pi }{2} - \frac{\pi }{6} = \frac{\pi }{3}\,\,{\text{or, }}{\cos ^{ - 1}}x = \frac{\pi }{6} \cr & \therefore \,\,x = \frac{{\sqrt 3 }}{2}. \cr} $$

$$\eqalign{ & \cot \left( {{\text{cose}}{{\text{c}}^{ - 1}}\frac{5}{3} + {{\tan }^{ - 1}}\frac{2}{3}} \right) \cr & = \cot \left[ {{{\tan }^{ - 1}}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}} \right] \cr & = \,\cot \,\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4} \times \frac{2}{3}}}} \right)} \right] \cr & = \,\cot \,\left[ {{{\tan }^{ - 1}}\frac{{17}}{6}} \right] \cr & = \,\cot \,\left( {{{\cot }^{ - 1}}\frac{6}{{17}}} \right) \cr & = \frac{6}{{17}} \cr} $$

$$\eqalign{ & {\text{Let, }}{S_\infty } = {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ..... \cr & \therefore {T_n} = {\cot ^{ - 1}}2{n^2} = {\tan ^{ - 1}}\frac{1}{{2{n^2}}} \cr & = {\tan ^{ - 1}}\left( {\frac{2}{{4{n^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\left( {2n + 1} \right) - \left( {2n - 1} \right)}}{{1 + \left( {2n + 1} \right)\left( {2n - 1} \right)}}} \right) \cr & = {\tan ^{ - 1}}\left( {2n + 1} \right) - {\tan ^{ - 1}}\left( {2n - 1} \right) \cr & \therefore {S_n} = \sum\limits_{n = 1}^\infty {\left\{ {{{\tan }^{ - 1}}\left( {2n + 1} \right) - {{\tan }^{ - 1}}\left( {2n - 1} \right)} \right\}} \cr & = {\tan ^{ - 1}}\infty - {\tan ^{ - 1}}1 = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4} \cr} $$

For $$f\left( x \right)$$  to be defined, we must have
$$\eqalign{ & - 1 \leqslant {\log _2}\left( {\frac{1}{2}{x^2}} \right) \leqslant 1 \cr & \Rightarrow {2^{ - 1}} \leqslant \frac{1}{2}{x^2} \leqslant {2^1} \cr & \Rightarrow 1 \leqslant {x^2} \leqslant 4\,\,\,.....\left( 1 \right) \cr & {\text{Now}},1 \leqslant {x^2} \cr & \Rightarrow {x^2} - 1 \geqslant 0{\text{ i}}{\text{.e}}{\text{. }}\left( {x - 1} \right)\left( {x + 1} \right) \geqslant 0 \cr & \Rightarrow x \leqslant - 1{\text{ or }}x \geqslant 1\,\,\,.....\left( 2 \right) \cr & {\text{Also}},\,\,{x^2} \leqslant 4 \cr & \Rightarrow {x^2} - 4 \leqslant 0{\text{ i}}{\text{.e}}{\text{. }}\left( {x - 2} \right)\left( {x + 2} \right) \leqslant 0 \cr & \Rightarrow - 2 \leqslant x \leqslant 2\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
Form (2) and (3), we get the domain of
$$\eqalign{ & f = \left\{ {\left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)} \right\} \cap \left[ { - 2,2} \right] \cr & = \left[ { - 2, - 1} \right] \cup \left[ {1,2} \right] \cr} $$

$$\eqalign{ & {\sin ^{ - 1}}\left( {x - 1} \right) \cr & \Rightarrow - 1 \leqslant x - 1 \leqslant 1 \cr & \Rightarrow 0 \leqslant x \leqslant 2 \cr & {\cos ^{ - 1}}\left( {x - 3} \right) \cr & \Rightarrow - 1 \leqslant x - 3 \leqslant 1 \cr & \Rightarrow 2 \leqslant x \leqslant 4 \cr & \therefore x = 2 \cr & {\text{So, }}{\sin ^{ - 1}}\left( {2 - 1} \right) + {\cos ^{ - 1}}\left( {2 - 3} \right) + {\tan ^{ - 1}}\frac{2}{{2 - 4}} = {\cos ^{ - 1}}k + \pi \cr & {\text{or, }}{\sin ^{ - 1}}1 + {\cos ^{ - 1}}\left( { - 1} \right) + {\tan ^{ - 1}}\left( { - 1} \right) = {\cos ^{ - 1}}k + \pi \cr & \Rightarrow \frac{\pi }{2} + \pi - \frac{\pi }{4} = {\cos ^{ - 1}}k + \pi \cr & \Rightarrow {\cos ^{ - 1}}k = \frac{\pi }{4}{\text{ or }}k = \frac{1}{{\sqrt 2 }} \cr} $$

$${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$
This is true only when $$x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - ..... = {x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4}.....$$
$$ \Rightarrow \frac{x}{{1 + \frac{x}{2}}} = \frac{{{x^2}}}{{1 + \frac{{{x^2}}}{2}}}$$
(Common ratios are $$ - \frac{x}{2}\,\,\& \,\, - \frac{{{x^2}}}{2}\,\,\& \,\,\left| {{\text{common ratios}}} \right| < 1,\,$$       in the given interval)
$$\eqalign{ & \frac{{2x}}{{2 + x}} = \frac{{2{x^2}}}{{2 + {x^2}}} \cr & \Rightarrow x = 0{\text{ or }}x = 1 \cr & \Rightarrow x = 1, \cr} $$
{ $$x$$ cannot be zero as $$0 < \left| x \right| < \sqrt 2 $$   }.

$$\eqalign{ & \frac{1}{{\sqrt 2 }}\left( {\cos\frac{{9\pi }}{{10}} - \sin \frac{{9\pi }}{{10}}} \right) = \cos \left( {\frac{\pi }{4} + \frac{{9\pi }}{{10}}} \right). \cr & \frac{1}{{\sqrt 2 }}\left( {\cos\frac{{9\pi }}{{10}} - \sin \frac{{9\pi }}{{10}}} \right) = \cos \frac{{23\pi }}{{20}} = \cos \left( {2\pi - \frac{{23\pi }}{{20}}} \right) = \cos \frac{{17\pi }}{{20}} \cr} $$
∴ the value $$ = {\cos ^{ - 1}}\left( {\cos \frac{{17\pi }}{{20}}} \right) = \frac{{17\pi }}{{20}}$$

$$\eqalign{ & {\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) \cr & \,\,\,\,\, = {\tan ^{ - 1}}x + 2{\tan ^{ - 1}}x = 3{\tan ^{ - 1}}x \cr & {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right] \cr & \Rightarrow \,\,y = \frac{{3x - {x^3}}}{{1 - 3{x^2}}} \cr} $$

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