સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
It is known that
$\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$
$\therefore$ $L.H.S.$ $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$
$=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$
$=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$
$=-2 \times(-\sin x)$
$=2 \sin x= R . H.S.$
Similar Questions
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
જો $\sin \alpha = \frac{{ - 3}}{5},$ કે જ્યાં $\pi < \alpha < \frac{{3\pi }}{2},$ તો $\cos \frac{1}{2}\alpha = $
જો $\sin \alpha = \frac{{ - 3}}{5},$ કે જ્યાં $\pi < \alpha < \frac{{3\pi }}{2},$ તો $\cos \frac{1}{2}\alpha = $
જો $A + B + C = {180^o},$ તો $\frac{{\sin 2A + \sin 2B + \sin 2C}}{{\cos A + \cos B + \cos C - 1}} = $
જો $A + B + C = {180^o},$ તો $\frac{{\sin 2A + \sin 2B + \sin 2C}}{{\cos A + \cos B + \cos C - 1}} = $
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
- [IIT 1993]
જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$
જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$