निम्नलिखित को सिद्ध कीजिए
$\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
$L.H.S$ $=\cot 4 x(\sin 5 x+\sin 3 x)$
$=\frac{\cot 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]$
$=2 \cos 4 x \cos x$
$R.H.S.$ $=\cot x(\sin 5 x-\sin 3 x)$
$=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$
$=2 \cos 4 x \cdot \cos x$
$L.H.S.$ $=$ $R.H.S.$
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
यदि $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} \right)$, तो $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}$ का मान होगा
निम्नलिखित को सिद्ध कीजिए
$\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
यदि $\alpha + \beta = \frac{\pi }{2}$ तथा $\beta + \gamma = \alpha ,$ तब $\tan \,\alpha $ =