સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}$
$=\frac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}$
$=\tan 4 x=R . H . S$
Similar Questions
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$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $
સમીકરણ ${\sin ^2}\,2\theta + {\cos ^4}\,2\theta = \frac{3}{4}$ ના $\theta \, \in \,\left( {0,\frac{\pi }{2}} \right)$ ના બધા ઉકેલો નો સરવાળો .......... થાય.
સમીકરણ ${\sin ^2}\,2\theta + {\cos ^4}\,2\theta = \frac{3}{4}$ ના $\theta \, \in \,\left( {0,\frac{\pi }{2}} \right)$ ના બધા ઉકેલો નો સરવાળો .......... થાય.
- [JEE MAIN 2019]
જો $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$ થાય તો
જો $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$ થાય તો
$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}} =$
$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}} =$
- [IIT 1985]
$cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ =
$cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ =