Inverse tangent in logarithmic-style algebraic form not used; standard algebraic identity for JEE

Formula

\(\tan^{-1}x=\sin^{-1}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)\)

Variables: \(x\in\mathbb{R}\).

Conditions: Principal values understood.

Changing inverse function type for simplification.