निम्नलिखित को सिद्ध कीजिए

$\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$

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It is known that 

$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$\therefore$ $L.H.S.$ $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$

$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$

$=\frac{\sin 2 x}{\cos 2 x}$

$=\tan 2 x$

$= R . H.S$

Similar Questions

$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $

$\tan 3A - \tan 2A - \tan A = $

${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $

यदि $\tan A = \frac{1}{2},$ तो $\tan 3A = $

$\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)$ का मान बराबर होगा।

  • [JEE MAIN 2022]