Question

Prove $\frac{1}{\sec x-\tan x}-\frac{1}{\cos x}=\frac{1}{\cos x}-\frac{1}{\sec x+\tan x}$

Answer 1

LHS

$\frac{1}{\sec x-\tan x}$$-\frac{1}{\cos x}$

$=\frac{1}{\frac{1}{\cos x}-\frac{\sin x}{\cos x}}-\frac{1}{\cos x}$

$=\frac{(\cos x{)}^2-1+\sin x}{\left(\cos x\right)(1-\sin x)}$

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

$=\frac{(\sin x{)}^2+\sin x}{\left(\cos x\right)(1-\sin x)}$

$=\frac{\left(\sin x\right)(1-\sin x)}{\left(\cos x\right)(1-\sin x)}=\tan x$

RHS

$=-\frac{1}{\sec x+\tan x}$$+\frac{1}{\cos x}$

$=-\frac{1}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}-\frac{1}{\cos x}$

$=\frac{-(\cos x{)}^2-+1+\sin x}{\left(\cos x\right)(1-\sin x)}$

$=\frac{(\sin x{)}^2+\sin x}{\left(\cos x\right)(1+\sin x)}$

$=\frac{\left(\sin x\right)(1+\sin x)}{\left(\cos x\right)(1+\sin x)}=\tan x$

LHS=RHS