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

Answer: 4

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