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If \( \sin \theta=\sin \alpha \), then
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Verified Answer
The correct answer is:
\( \frac{\theta+\alpha}{2} \) is any odd multiple of \( \frac{\Pi}{2} \) and \( \frac{\theta-\alpha}{2} \) is any multiple of
Given $\sin \theta=\sin \alpha$
$\sin \theta-\sin \alpha=0$
$2 \cos \left(\frac{\theta+\alpha}{2}\right) \sin \left(\frac{\theta-\alpha}{2}\right)=0$
It is not necessary that $\frac{\theta+\alpha}{2}$ is odd multiple of $\pi / 2$ and $\frac{\theta-\alpha}{2}$ is any multiple of $\pi$ should be simultaneous hold good for
the above equation to be true
Hence, the correct answer should be
$\frac{\theta+\alpha}{2}$ is any odd multiple of $\pi / 2$
or $\frac{\theta-\alpha}{2}$ is any multiple of $\pi$
$\sin \theta-\sin \alpha=0$
$2 \cos \left(\frac{\theta+\alpha}{2}\right) \sin \left(\frac{\theta-\alpha}{2}\right)=0$
It is not necessary that $\frac{\theta+\alpha}{2}$ is odd multiple of $\pi / 2$ and $\frac{\theta-\alpha}{2}$ is any multiple of $\pi$ should be simultaneous hold good for
the above equation to be true
Hence, the correct answer should be
$\frac{\theta+\alpha}{2}$ is any odd multiple of $\pi / 2$
or $\frac{\theta-\alpha}{2}$ is any multiple of $\pi$
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