Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We have $\cos \left({\cot }^{-1}\left(\frac{1}{2}\right)\right)=\cot \left({\cos }^{-1}x\right)$ Let ${\cot }^{-1}\left(\frac{1}{2}\right)=\alpha \Rightarrow \cot \alpha =\frac{1}{2}\Rightarrow {\cos }^{-1}\frac{1}{\sqrt{5}}$ $\Rightarrow \cos \left({\cos }^{-1}\frac{1}{\sqrt{5}}\right)=\cot \left({\cot }^{-1}\frac{x}{\sqrt{1-x^2}}\right)$ $\Rightarrow \frac{1}{\sqrt{5}}=\frac{x}{\sqrt{1-x^2}}\Rightarrow \sqrt{1-x^2}=\sqrt{5}x$ On squaring both sides, we get, $1-x^2=5x^2\Rightarrow 1=6x^2$ $\Rightarrow x=\pm \frac{1}{\sqrt{6}}$ $\begin{array}{ll}& x=\frac{1}{\sqrt{6}}\text{ [neglecting - ve sign] }\end{array}$