જો $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, તો $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = . . .$
જો $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, તો $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = . . .$
- A
$\frac{1}{2}$
- B
$1\over3$
- C
$\frac{1}{4}$
- D
$\frac{1}{8}$
Similar Questions
જો $A = 133^\circ ,$ તો $\;2\cos \frac{A}{2} = . . . .$
જો $A = 133^\circ ,$ તો $\;2\cos \frac{A}{2} = . . . .$
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
- [IIT 1977]
જો $\theta $ એ લઘુકોણ છે અને $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, તો $\tan \theta = . . .$
જો $\theta $ એ લઘુકોણ છે અને $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, તો $\tan \theta = . . .$
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$\tan 5x\tan 3x\tan 2x = $
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
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