$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
- A
$\frac{1}{2}\tan \theta $
- B
$\frac{1}{2}\cot \theta $
- C
$\tan \theta $
- D
$\cot \theta $
Similar Questions
$[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ =
$[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ =
જો $a{\sin ^2}x + b{\cos ^2}x = c,\,\,$$b\,{\sin ^2}y + a\,{\cos ^2}y = d$ અને $a\,\tan x = b\,\tan y,$ તો $\frac{{{a^2}}}{{{b^2}}} = . . ..$
જો $a{\sin ^2}x + b{\cos ^2}x = c,\,\,$$b\,{\sin ^2}y + a\,{\cos ^2}y = d$ અને $a\,\tan x = b\,\tan y,$ તો $\frac{{{a^2}}}{{{b^2}}} = . . ..$
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
જો $\sin \alpha = \frac{{336}}{{625}}$ અને $450^\circ < \alpha < 540^\circ ,$ તો $\sin \left( {\frac{\alpha }{4}} \right) = $
જો $\sin \alpha = \frac{{336}}{{625}}$ અને $450^\circ < \alpha < 540^\circ ,$ તો $\sin \left( {\frac{\alpha }{4}} \right) = $
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
- [IIT 1979]