Question

If $\cot (\alpha +\beta )=0,$ then $\sin (\alpha +2\beta )$ is equal to

• A. $\sin \alpha$
• B. $\sin 2\beta$
• C. $\cos \alpha$
• D. $\sin \beta$

Answer 1

The correct option is A $\sin \alpha$

$\cot (\alpha +\beta )=0\Rightarrow \cos (\alpha +\beta )=0\Rightarrow \cos \alpha \cos \beta -\sin \alpha \sin \beta =0\Rightarrow \cos \alpha \cos \beta =\sin \alpha \sin \beta$
Now $\sin (\alpha +2\beta )=\sin (\alpha +\beta +\beta )=\sin (\alpha +\beta )\cos \beta +\cos (\alpha +\beta )\sin \beta =\sin (\alpha +\beta )\cos \beta [since,cos(\alpha +\beta )=0]=(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\cos \beta =\sin \alpha {\cos }^2\beta +\sin \beta .\cos \alpha \cos \beta =\sin \alpha {\cos }^2\beta +\sin \beta .\sin \alpha \sin \beta =\sin \alpha [{\cos }^2\beta +{\sin }^2\beta ]=\sin \alpha \left(1\right)=\sin \alpha$

Therefore, Answer is $\sin \alpha$