Question

Prove that
$\frac{\sin 9\theta -\sin 7\theta }{\cos 7\theta -\cos 9\theta }=\cot 8\theta$.

Answer 1

L.H.S $=\frac{\sin 9\theta -\sin 7\theta }{\cos 7\theta -\cos 9\theta }=\cot 8\theta$

$=\left(\frac{2\cos \left(\frac{16\theta }{2}\right)\sin \left(\frac{2\theta }{2}\right)}{-2\sin \left(\frac{9\theta +7\theta }{2}\right)\sin \left(\frac{7\theta -9\theta }{2}\right)}\right)$ ....... (Using $\sin a-\sin b$)

$=\left(\frac{2\cos 8\theta \sin \theta }{2\sin 8\theta \sin \theta }\right)$

$=\cot 8\theta$

$=$ R.H.S