Question

If $\frac{3-\tan ^{2} \frac{\pi}{7}}{1-\tan ^{2} \frac{\pi}{7}}= K \cos \frac{\pi}{7}$ then $K$ equals.

Answer 1

$ LHS =\frac{3 \cos ^{2} \frac{\pi}{7}-\sin ^{2} \frac{\pi}{7}}{\cos ^{2} \frac{\pi}{7}-\sin ^{2} \frac{\pi}{7}}=\frac{4 \cos ^{2} \frac{\pi}{7}-1}{\cos \frac{2 \pi}{7}} $
$=\frac{1+2 \cos \frac{2 \pi}{7}}{\cos \frac{2 \pi}{7}}=\frac{2\left[\sin \frac{2 \pi}{7}+\sin \frac{4 \pi}{7}\right]}{\sin \frac{4 \pi}{7}}$
(Multiplying Nr & Dr by $2 \sin \frac{2 \pi}{7}$ )
$=\frac{4 \sin \frac{3 \pi}{7} \cdot \cos \frac{\pi}{7}}{\sin \frac{4 \pi}{7}}=4 \cos \frac{\pi}{7}$
$\left(\because \sin \frac{3 \pi}{7}=\sin \frac{4 \pi}{7}\right)$