If operatorname{cosec}^2(alpha+beta)-sin ^2(b | vedclass

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Question

(a)

$\operatorname{cosec}^2(\alpha+\beta)-\sin ^2(\beta-\alpha)$

$+\sin ^2(2 \alpha-\beta)=\cos ^2(\alpha-\beta)$

$\Rightarrow \operatorname{

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(a)

$\operatorname{cosec}^2(\alpha+\beta)-\sin ^2(\beta-\alpha)$

$+\sin ^2(2 \alpha-\beta)=\cos ^2(\alpha-\beta)$

$\Rightarrow \operatorname{cosec}^2(\alpha+\beta)+\sin ^2(2 \alpha-\beta)$

$=\cos ^2(\alpha-\beta)+\sin ^2(\alpha-\beta)$

$\Rightarrow \operatorname{cosec}^2(\alpha+\beta)+\sin ^2(2 \alpha-\beta)=1$

lt is possible only $\operatorname{cosec}^2(\alpha+\beta)=1$ and $\sin ^2(2 \alpha-\beta)=0$

$	herefore \quad \alpha+\beta=\frac{\pi}{2}$ and $2 \alpha-\beta=0$

On solving these equations, we get

$\alpha=\frac{\pi}{6} 	ext { and } \beta=\frac{\pi}{3}$

$	herefore \sin (\alpha-\beta)=\sin \left(\frac{\pi}{6}-\frac{\pi}{3}ight)=\sin \left(-\frac{\pi}{6}ight)=-\frac{1}{2}$

If $\operatorname{cosec}^2(\alpha+\beta)-\sin ^2(\beta-\alpha)+\sin ^2(2 \alpha-\beta)=\cos ^2(\alpha-\beta)$ where $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\sin (\alpha-\beta)$ is equal to

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