Prove that:
$ 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
$L.H.S.$ $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$
$=2 \cos \frac{\pi}{3} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)$
$\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \left(\frac{-\pi}{13}\right)$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$
$=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13}$
$=2 \cos \frac{\pi}{13}\left[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}\right]$
$=2 \cos \frac{\pi}{13}\left[2 \cos \left(\frac{\frac{9 \pi}{13}+\frac{4 \pi}{13}}{2}\right) \cos \frac{\frac{9 \pi}{13}-\frac{4 \pi}{13}}{2}\right]$
$=2 \cos \frac{\pi}{13}\left[2 \cos \frac{\pi}{2} \cos \frac{5 \pi}{26}\right]$
$=2 \cos \frac{\pi}{13} \times 2 \times 0 \times \cos \frac{5 \pi}{26}$
$=0=R . H . S.$
If $\sec \theta + \tan \theta = p,$ then $\tan \theta $ is equal to
If $p = \frac{{2\sin \,\theta }}{{1 + \cos \theta + \sin \theta }}$, and $q = \frac{{\cos \theta }}{{1 + \sin \theta }},$ then
Let $S_1,S_2$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangent to each pair of circles are drawn and extended so that they can intersect and form a triangle $ABC$ with circumradius $R,$ then $R$ is equal to
Find the degree measures corresponding to the following radian measures (Use $\pi=\frac{22}{7}$ ).
$-4$
The value of the expression $1 - \frac{{{{\sin }^2}y}}{{1 + \cos \,y}} + \frac{{1 + \cos \,y}}{{\sin \,y}} - \frac{{\sin \,\,y}}{{1 - \cos \,y}}$ is equal to