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If $\cot (A+B)=0$, then $\sin (A+2 B)$ is equal to
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Verified Answer
The correct answer is:
$\sin A$
$\cot (A+B)=0 \Rightarrow A+B=(2 n+1) \frac{\pi}{2} \Rightarrow B=(2 n+1) \frac{\pi}{2}-A$
Now, $\sin (A+2 B)=\sin (A+(2 n+1) \pi-2 A)$
$=\sin ((2 n+1) \pi-A)=\sin A$
Now, $\sin (A+2 B)=\sin (A+(2 n+1) \pi-2 A)$
$=\sin ((2 n+1) \pi-A)=\sin A$
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