સાબિત કરો કે : $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
$L.H.S$ $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$
$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
$= R . H.S$
જો $\alpha = 22^\circ 30' $ તો $(1 + \cos \alpha )(1 + \cos 3\alpha )$ $(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ = . . .. .
જો $\tan \theta = \frac{{20}}{{21}},$ cos$\theta$ મેળવો.
સાબિત કરો કે : $ 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
$1 - \frac{{{{\sin }^2}y}}{{1 + \cos \,y}} + \frac{{1 + \cos \,y}}{{\sin \,y}} - \frac{{\sin \,\,y}}{{1 - \cos \,y}} =$
જો $\sin \theta + \cos \theta = m$ અને $\sec \theta + {\rm{cosec}}\theta = n$, તો $n(m + 1)(m - 1) = $