સાબિત કરો કે : $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
સાબિત કરો કે : $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\sin ^{2} 6 x-\sin ^{2} 4 x$
$=(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x)$
$=\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \cdot \sin \left(\frac{6 x-4 x}{2}\right)\right]$
$=(2 \sin 5 x \cos x)(2 \cos 5 x \sin x)=(2 \sin 5 x \cos 5 x)(2 \sin x \cos x)$
$=\sin 10 x \sin 2 x$
$= R . H.S$
Similar Questions
સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
- [IIT 1986]
જો $\sin A = n\sin B,$ તો $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
જો $\sin A = n\sin B,$ તો $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $