સાબિત કરો કે : $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

It is known that

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

$L.H.S.$ $=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$

$=\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left[2 \cos \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \cos \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}$

$=\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 6 x]}$

$=\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]}$

$=\tan 6 x$

$= R . H.S.$

Similar Questions

$37.4$ સેમી ચાપની લંબાઈ ધરાવતા તથા કેન્દ્ર આગળ $60^{\circ}$ માપનો ખૂણો બનાવતા વર્તુળની ત્રિજ્યા શોધો. ( $\pi=\frac{22}{7}$ લો ).

જો  $\tan \theta = \frac{{x\,\sin \,\phi }}{{1 - x\,\cos \,\phi }}$ અને  $\tan \,\phi  = \frac{{y\sin \,\theta }}{{1 - y\,\cos \,\theta }}$, તો  $\frac{x}{y} = $

મૂલ્ય શોધો. $\cot \left(-\frac{15 \pi}{4}\right)$

$\cos 15^\circ = $

જો  $x = \sec \theta + \tan \theta ,$ તો  $x + \frac{1}{x} = $