If cot (α + β) = 0, then write the value of sin (α + 2β).

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#### Solution

\[\cot \left( \alpha + \beta \right) = 0\]

\[ \Rightarrow \alpha + \beta = \frac{\pi}{2} \left( 1 \right)\]

\[\beta = \frac{\pi}{2} - \alpha \left( 2 \right)\]

\[\alpha = \frac{\pi}{2} - \beta \left( 3 \right)\]

\[\text{ Now, }\sin\left( \alpha + 2\beta \right) = \sin\left( \alpha + \beta + \beta \right)\]

\[ = \sin\left( \frac{\pi}{2} + \frac{\pi}{2} - \alpha \right)\]

\[ = \sin\left( \pi - \alpha \right)\]

\[ = \sin \alpha\]

\[\text{ Now, }\sin\left( \alpha + 2\beta \right) = \sin\left( \alpha + 2\beta \right)\]

\[ = \sin\left( \frac{\pi}{2} - \beta + 2\beta \right)\]

\[ = \sin\left( \frac{\pi}{2} + \beta \right)\]

\[ = \cos \beta\]

Is there an error in this question or solution?

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