સાબિત કરો કે : $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
$L.H.S.$ $=\sin 3 x+\sin 2 x-\sin x$
$=\sin 3 x+\left[2 \cos \left(\frac{2 x+x}{2}\right) \sin \left(\frac{2 x-x}{2}\right)\right]$
$\left[\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\sin 3 x+\left[2 \cos \left(\frac{3 x}{2}\right) \sin \left(\frac{x}{2}\right)\right]$
$=\sin 3 x+2 \cos \frac{3 x}{2} \sin \frac{x}{2}$
$=2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2}+2 \cos \frac{3 x}{2} \sin \frac{x}{2}$$\quad[\sin 2 A=2 \sin A \cdot \cos B]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[\sin \left(\frac{3 x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[2 \sin \left\{\frac{\left(\frac{3 x}{2}\right)+\left(\frac{x}{2}\right)}{2}\right\} \cos \left\{\frac{\left(\frac{3 x}{2}\right)-\left(\frac{x}{2}\right)}{2}\right\}\right]$
$\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right) \cdot 2 \sin x \cos \left(\frac{x}{2}\right)$
$=4 \sin x \cos \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)= R. H.S.$
સાબિત કરો કે : $ 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
જો $\sin \theta + {\rm{cosec}}\theta = 2,$ તો ${\sin ^{10}}\theta + {\rm{cose}}{{\rm{c}}^{10}}\theta $ = . . .
જો $\sin x=-\frac{3}{5}$, જ્યાં $\pi < x < \frac{3 \pi}{2}$, તો $80\left(\tan ^2 x-\cos x\right)=$...........
સાબિત કરો કે : $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$
મૂલ્ય શોધો. $\cot \left(-\frac{15 \pi}{4}\right)$