MCQ
જો $\cot (\alpha + \beta ) = 0,$ તો $\sin (\alpha + 2\beta ) = $
- ✓$\sin \alpha $
- B$\cos \alpha $
- C$\sin \beta $
- D$\cos 2\beta $
==> $\alpha + \beta = (2n + 1)\frac{\pi }{2},n \in I$
$\therefore$ $\sin (\alpha + 2\beta ) = \sin (2\alpha + 2\beta - \alpha )$
$=\sin {\rm{ }}[(2n + 1){\rm{ }}\pi - \alpha ]$
$ = \sin (\,2n\pi + \pi - \alpha \,)$ = $\sin (\,\pi - \alpha \,)\, = \sin \alpha $.
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