Prove the following identity:

$\sqrt{\sec^2 A + \cosec^2 A} = \tan A + \cot A$

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\sec^2 A + \cosec^2 A} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\cos^2 A} + \dfrac{1}{\sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\cos^2 A \sin^2 A}} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A}.$

Solving R.H.S. of the equation :

$\Rightarrow \tan A + \cot A \\[1em] \Rightarrow \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\sec^2 A + \cosec^2 A} = \tan A + \cot A$.