2. inverse trigonometric function — Maths STD 12 Science — Question

Let ${\cot ^{ - 1}}\frac{1}{2} = \phi \Rightarrow \frac{1}{2} = \cot \phi $

$ \Rightarrow \sin \phi = \frac{1}{{\sqrt {1 + {{\cot }^2}\phi } }} = \frac{2}{{\sqrt 5 }}$

Let ${\cos ^{ - 1}}x = \theta \Rightarrow \sec \theta = \frac{1}{x} $

$\Rightarrow \tan \theta = \sqrt {{{\sec }^2}\theta - 1} $

$ \Rightarrow \tan \theta = \sqrt {\frac{1}{{{x^2}}} - 1} $

$\Rightarrow \tan \theta = \frac{{\sqrt {1 - {x^2}} }}{x}$

So, $\tan \{ {\cos ^{ - 1}}(x)\} = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right)$

$ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = \sin \left( {{{\sin }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$

$ \Rightarrow \frac{{\sqrt {1 - {x^2}} }}{x} = \frac{2}{{\sqrt 5 }} \Rightarrow \sqrt {(1 - {x^2})5} = 2x$

Squaring both sides, we get $x = \pm \frac{{\sqrt 5 }}{3}$.