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(A) Given,$\cot (\alpha + \beta ) = 0.$Since $\cot 	heta = 0$ implies $	heta = (2n + 1)\frac{\pi}{2}$ for $n \in I,$we have $\alpha + \beta = (2n +

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$\sin \alpha $

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(A) Given,$\cot (\alpha + \beta ) = 0.$Since $\cot 	heta = 0$ implies $	heta = (2n + 1)\frac{\pi}{2}$ for $n \in I,$we have $\alpha + \beta = (2n + 1)\frac{\pi}{2}.$Now,consider $\sin (\alpha + 2\beta ) = \sin ((\alpha + \beta ) + \beta ).$Substituting $\alpha + \beta = (2n + 1)\frac{\pi}{2},$$\sin (\alpha + 2\beta ) = \sin ((2n + 1)\frac{\pi}{2} + \beta ) = \cos \beta$ if $n$ is even,or $-\cos \beta$ if $n$ is odd.Alternatively,using the identity $\sin (\alpha + 2\beta ) = \sin (2(\alpha + \beta ) - \alpha ) = \sin ((2n + 1)\pi - \alpha ).$Since $\sin ((2n + 1)\pi - \alpha ) = \sin (\pi - \alpha ) = \sin \alpha$ for even $n,$ and $\sin (3\pi - \alpha ) = \sin \alpha,$ the result is $\sin \alpha.$

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

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