Question

Prove that
$\frac{\cos 2B+\cos 2A}{\cos 2B-\cos 2A}=\cot (A+B)\cot (A-B)$.

Answer 1

L.H.S$=\frac{\cos 2B+\cos 2A}{\cos 2B-\cos 2A}$

$=\frac{2\cos \left(\frac{2B+2A}{2}\right)\cos \left(\frac{2B-2A}{2}\right)}{-2\sin \left(\frac{2B+2A}{2}\right)\sin \left(\frac{2B-2A}{2}\right)}$ using transformation angle formula, $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$ and $\cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$=\frac{\cos (A+B)\cos (A-B)}{\sin (A+B)\sin (A-B)}$ using $\sin (-\theta )=-\sin \theta$ and $\cos (-\theta )=\cos \theta$

$=\frac{\cos (A+B)\cos (A-B)}{\sin (A+B)\sin (A-B)}$

$=\cot (A+B)\cot (A-B)=$R.H.S

Hence proved.