Question

Prove the following identity:
iii.$\frac{\sin A+\cos A}{\sin A–\cos A}+\frac{\sin A–\cos A}{sinA+\cos A}=\frac{2}{(2{\sin }^{2}A–1)}$

Answer 1

Proving LHS $=$ RHS.

First we consider Left Hand Side (LHS),

$=\frac{\sin A+\cos A}{\sin A–\cos A}+\frac{\sin A–\cos A}{\sin A+\cos A}$ .

By taking LCM we get,

$=\frac{(\sin A+cosA)2+(\sin A–\cos A)2}{(\sin A+\cos A)(\sin A–cosA)}$

$=\frac{{\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A+{\sin }^{2}A+{\cos }^{2}A–2\sin \cos A}{({\sin }^{2}A–co{s}^{2}A)}$

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

$=\frac{2({\sin }^{2}A+co{s}^{2}A)}{(si{n}^{2}A–{\cos }^{2}A)}$

We know that, ${\sin }^{2}A+{\cos }^{2}A=1$

$=\frac{2}{({\sin }^{2}A–{\cos }^{2}A)}$

$=\frac{2}{[{\sin }^{2}A–(1–{\sin }^{2}A)]}$

$=\frac{2}{(2{\sin }^{2}A-1)}$

Then, Right Hand Side $=\frac{2}{(2{\sin }^{2}A-1)}$

Therefore, LHS = RHS.

Hence proved.