Question

Prove (i) $\tan x+\cot x=\sec x\text{cosec}x$
(ii) ${\sec }^{2}x+{\text{cosec}}^{2}x={\sec }^{2}x\cdot {\mathrm{cosec}}^{2}x$

Answer 1

Consider the L.H.S.

$=\tan x+\cot x$

$=\frac{\sin x}{\cos x}+\frac{cosx}{sinx}$

$=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x.\cos x}$

We know that

${\sin }^{2}x+{\cos }^{2}x=1$

Therefore,

$=\frac{1}{\sin x.cosx}$

$=\sec x.\text{cosec}x$

Hence, proved.

We have,

$={\sec }^{2}x+{\mathrm{cosec}}^{2}x$

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

$=\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}$

$=\frac{1}{{\sin }^{2}x.{\cos }^{2}x}$

$={\sec }^{2}x.{\mathrm{cosec}}^{2}x$

Hence, proved.