सिद्ध कीजिए
$2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
$L.H.S.$ $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$
$=2 \cos \frac{\pi}{3} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)$
$\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \left(\frac{-\pi}{13}\right)$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$
$=2 \cos \frac{\pi}{13}\left[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}\right]$
$=2 \cos \frac{\pi}{13}\left[2 \cos \left(\frac{\frac{9 \pi}{13}+\frac{4 \pi}{13}}{2}\right) \cos \frac{\frac{9 \pi}{13}-\frac{4 \pi}{13}}{2}\right]$
$=2 \cos \frac{\pi}{13}\left[2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}\right]$
$=2 \cos \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26}$
$=0=R . H . S.$
समीकरण ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ तभी सम्भव है जब
यदि $x\sin 45^\circ {\cos ^2}60^\circ = \frac{{{{\tan }^2}60^\circ {\rm{cosec}}30^\circ }}{{\sec 45^\circ {{\cot }^2}30^\circ }},$ तब $x = $
यदि $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ तो
$2 \sin \left(12^{\circ}\right)-\sin \left(72^{\circ}\right)$ का मान होगा-
यदि $\tan \theta + \sin \theta = m$ तथा $\tan \theta - \sin \theta = n,$ तो