Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

Given that, $\cot \left({\cos }^{-1}x\right)=\sec \left\{{\tan }^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}$ Since, ${\cos }^{-1}x={\cot }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and ${\tan }^{-1}x={\sec }^{-1}\left(\sqrt{1+x^2}\right)$ $\Rightarrow \cot \left({\cot }^{-1}\frac{x}{\sqrt{1-x^2}}\right)=$ $\sec \left\{{\sec }^{-1}\sqrt{1+{\left(\frac{a}{\sqrt{b^2-a^2}}\right)}^2}\right\}$ $\Rightarrow \frac{x}{\sqrt{1-x^2}}=\sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}}$ $\Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}}$ On squaring both the sides, we get $\Rightarrow \frac{x^2}{1-x^2}=\frac{b^2}{b^2-a^2}$ $\begin{array}{rl}\Rightarrow & x^2{b}^2-x^2{a}^2=b^2-b^2{x}^2\\ & x^2{b}^2+b^2{x}^2-x^2{a}^2=b^2\end{array}$ $\Rightarrow 2x^2{b}^2-x^2{a}^2=b^2\Rightarrow x^2\left(2b^2-a^2\right)=b^2$ $\Rightarrow x=\frac{b}{\sqrt{2b^2-a^2}}$