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

Solving L.H.S. of the above equation :

$\Rightarrow \sqrt{\Big(\dfrac{1}{\text{cos A}}\Big)^2 + \Big(\dfrac{1}{\text{sin A}}\Big)^2} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A\text{ sin}^2 A}}$

By formula,

sin² A + cos² A = 1

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

Solving R.H.S. of the equation :

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

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

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