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Trigonometrical Equations
hard
જો $\cot (\alpha + \beta ) = 0,$ તો $\sin (\alpha + 2\beta ) = $
A
$\sin \alpha $
B
$\cos \alpha $
C
$\sin \beta $
D
$\cos 2\beta $
Solution
(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 $.
Standard 11
Mathematics
