If 2 tan

$\mathrm{RHS}=\frac{3+5 \cos \beta}{5+3 \cos \beta}$

$=\frac{3+5\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}{5+3\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}$

$=\frac{3+3 \tan ^{2} \frac{\beta}{2}+5-5 \tan ^{2} \frac{\beta}{2}}{5+5 \tan ^{2} \frac{\beta}{2}+3-3 \tan \frac{\beta}{2}}$

$=\frac{8-2 \tan ^{2} \frac{\beta}{2}}{8+2 \tan ^{2} \frac{\beta}{2}}$

$=\frac{8-8 \tan ^{2} \frac{\alpha}{2}}{8+8 \tan ^{2} \frac{\alpha}{2}}$            $\left[\because 2 \tan \frac{\alpha}{2}=\tan \frac{\beta}{2}\right]$

$=\frac{8\left(1-\tan ^{2} \frac{\alpha}{2}\right)}{8\left(1+\tan ^{2} \frac{\alpha}{2}\right)}$

$=\frac{1-\tan ^{2} \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}$

$=\cos \alpha=\mathrm{LHS}$

Hence proved.