જો $A + B + C = {180^o},$ તો $\frac{{\sin 2A + \sin 2B + \sin 2C}}{{\cos A + \cos B + \cos C - 1}} = $
જો $A + B + C = {180^o},$ તો $\frac{{\sin 2A + \sin 2B + \sin 2C}}{{\cos A + \cos B + \cos C - 1}} = $
- A
$8\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
- B
$8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- C
$8\,\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- D
$8\,\cos \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
Similar Questions
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જો $\sin \alpha = \frac{{336}}{{625}}$ અને $450^\circ < \alpha < 540^\circ ,$ તો $\sin \left( {\frac{\alpha }{4}} \right) = $
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
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${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8}$ =
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જો $\alpha $ સમીકરણ $25{\cos ^2}\theta + 5\cos \theta - 12 = 0$, $\pi /2 < \alpha < \pi $, નું એક બીજ હોય તો $\sin 2\alpha = . . .$
જો $\alpha $ સમીકરણ $25{\cos ^2}\theta + 5\cos \theta - 12 = 0$, $\pi /2 < \alpha < \pi $, નું એક બીજ હોય તો $\sin 2\alpha = . . .$