Question

SOLVE THE ABOVE problem​

Answer 1

$cos(\theta+2\alpha)=mcos\theta$

or, $\frac{cos(\theta+2\alpha)}{cos\theta}=\frac{m}{1}$

use Componendo and Dividendo rule,

$\frac{cos(\theta+2\alpha)+cos\theta}{cos\theta-cos(\theta+2\alpha)}=\frac{m+1}{1-m}$

use formula, cosA + cosB = 2cos(A + B//2.cos(A - B)/2

and cosA - cosB = 2sin(A + B)/2 sin(B - A)/2

$\frac{2cos(\theta+\alpha).cos\alpha}{2sin(\theta+\alpha).sin\alpha}=\frac{m+1}{1-m}$

$\frac{cos(\theta+\alpha)}{sin(\theta+\alpha)}\frac{cos\alpha}{sin\alpha}=\frac{m+1}{1-m}$

$cot(\theta+\alpha).cot\alpha=\frac{m+1}{1-m}$

$\frac{cot\alpha}{tan(\theta+\alpha)}=\frac{m+1}{1-m}$

$(1-m)cot\alpha=(m+1)tan(\theta+\alpha)$

hence proved.