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

\[\begin{aligned} \text { L.H.S. } & =\frac{\sec ^{2} x-\tan ^{2} x}{\sec x-\tan x}-\sec x \\ & =\sec x+\tan x-\sec x \\ & =\tan x \end{aligned} \] \[\begin{aligned} \text { R.H.S. } & =\sec x-\frac{\sec ^{2} x-\tan ^{2} x}{\sec x+\tan x} \\ & =\sec x-\sec x+\tan x \\ & =\tan x \\ \text { LHS } & =\text { RHS } \end{aligned} \] Alternate Solution: Reframing, \(\displaystyle \frac{1}{\sec x-\tan x}+\frac{1}{\sec x+\tan x}=\frac{2}{\cos x}\) \[\begin{aligned} \mathrm{LHS} & =\frac{(\sec x+\tan x)+(\sec x-\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} \\ & =\frac{2 \sec x}{\sec ^{2} x-\tan ^{2} \mathrm{x}} \\ & =2 \sec x \\ & =\frac{2}{\cos x}=\mathrm{RHS} \end{aligned} \]