If cot(α + β) = 0, Find sin(α + 2β)

Solution

Given,

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

We know that

\[ \cot\theta=0 \]

when

\[ \theta=\frac{\pi}{2} \]

Therefore,

\[ \alpha+\beta=\frac{\pi}{2} \]

Now,

\[ \sin(\alpha+2\beta) = \sin[(\alpha+\beta)+\beta] \]

Substituting

\[ \alpha+\beta=\frac{\pi}{2} \]

we get

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

Using the identity:

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

Therefore,

\[ \sin(\alpha+2\beta)=\cos\beta \]

Now from

\[ \alpha+\beta=\frac{\pi}{2} \]

we get

\[ \beta=\frac{\pi}{2}-\alpha \]

Hence,

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

Therefore,

\[ \boxed{ \sin(\alpha+2\beta)=\sin\alpha } \]

Final Answer

\[ \boxed{ \sin(\alpha+2\beta)=\sin\alpha } \]

Correct Option: (a)