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
$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ =
$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ =
- [JEE MAIN 2022]
જો $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ હોય તો,
જો $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ હોય તો,
$\tan 7\frac{1}{2}^\circ =...$
$\tan 7\frac{1}{2}^\circ =...$
સાબિત કરો કે : $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
સાબિત કરો કે : $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
જો $A + B + C = {180^o},$ તો $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B) = . . . .$
જો $A + B + C = {180^o},$ તો $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B) = . . . .$