If \(\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)\), then: [JEE Main…

\(\text{ Given, }\cot \left({\cos }^{-1}x\right)=\sec \left({\tan }^{-1}\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow cot\left({\cot }^{-1}\left(\frac{x}{\sqrt{1-{x}^{2}}}\right)\right)=\sec \left({\sec }^{-1}\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\right)\\ \Rightarrow \frac{x}{\sqrt{1-{x}^{2}}}=\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}\\ \Rightarrow \frac{b}{\sqrt{2{b}^{2}-{a}^{2}}}=x\)