MCQ
If $\cot (\alpha + \beta ) = 0,$ then $\sin (\alpha + 2\beta ) = $
  • $\sin \alpha $
  • B
    $\cos \alpha $
  • C
    $\sin \beta $
  • D
    $\cos 2\beta $

Answer

Correct option: A.
$\sin \alpha $
a
(a) Given, $\cot (\alpha + \beta ) = 0 \Rightarrow \cos (\alpha + \beta ) = 0$

==> $\alpha + \beta = (2n + 1)\frac{\pi }{2},n \in I$

$\therefore$ $\sin (\alpha + 2\beta ) = \sin (2\alpha + 2\beta - \alpha )$

$=\sin {\rm{ }}[(2n + 1){\rm{ }}\pi - \alpha ]$

$ = \sin (\,2n\pi + \pi - \alpha \,)$ = $\sin (\,\pi - \alpha \,)\, = \sin \alpha $.

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