Question

Prove that: $\sqrt{{\sec }^{2}A+{\csc }^{2}A}=\tan A+\cot A$

Answer 1

$LHS$

$=\sqrt{{\sec }^{2}A+{\csc }^{2}A}$

$=\sqrt{\frac{1}{{\cos }^{2}A}+\frac{1}{{\sin }^{2}A}}$

$=\sqrt{\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}}$

$=\sqrt{\frac{1}{{\sin }^{2}A{\cos }^{2}A}}$

$=\csc A\sec A.$

Now,

$RHS$

$=\tan A+\cot A$

$=\frac{\sin A}{coA}+\frac{\cos A}{\sin A}$

$=\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}$

$==\frac{1}{\sin A\cos A}$

$=\csc A\sec A$

$=LHS.$