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}$$\quad[\sin 2 A=2 \sin A \cdot \cos B]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[\sin \left(\frac{3 x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[2 \sin \left\{\frac{\left(\frac{3 x}{2}\right)+\left(\frac{x}{2}\right)}{2}\right\} \cos \left\{\frac{\left(\frac{3 x}{2}\right)-\left(\frac{x}{2}\right)}{2}\right\}\right]$
$\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right) \cdot 2 \sin x \cos \left(\frac{x}{2}\right)$
$=4 \sin x \cos \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)= R. H.S.$
જો $5\tan \theta = 4,$ તો $\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = $
જો સમીકરણ ${\sin ^2}\theta = \frac{{{x^2} + {y^2}}}{{2xy}},x,y, \ne 0$ શકય હોય તો
$\tan \frac{13 \pi}{12}$ નું મૂલ્ય શોધો.
રેડિયન માપ શોધો : $-47^{\circ} 30^{\prime}$
જો $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$
$ = \tan \alpha \tan \beta \tan \gamma $, તો $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $