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

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Question

(A) Given the equation: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.Rearranging the terms,w

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$-\frac{1}{2}$

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(A) Given the equation: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.Rearranging the terms,we get: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\beta-\alpha)$.Since $\sin^2(\beta-\alpha) = \sin^2(\alpha-\beta)$,the equation becomes: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)$.Using the identity $\cos^2	heta+\sin^2	heta=1$,we have: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=1$.Since $\operatorname{cosec}^2(\alpha+\beta) \geq 1$ and $\sin^2(2\alpha-\beta) \geq 0$,the sum can be $1$ only if $\operatorname{cosec}^2(\alpha+\beta)=1$ and $\sin^2(2\alpha-\beta)=0$.This implies $\alpha+\beta = \frac{\pi}{2}$ and $2\alpha-\beta = 0$.Adding the two equations: $3\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{6}$.Substituting $\alpha$ back: $\beta = 2\alpha = \frac{\pi}{3}$.Thus,$\sin(\alpha-\beta) = \sin(\frac{\pi}{6}-\frac{\pi}{3}) = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$.

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

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