If cot⁡(α+β)=0, then sin⁡(α+2β) can be

A
$- \sin α$
B
$\sin β$
C
$\cos α$
D
$\cos β$

Solution:

$\begin{array}{l} \cot (α + β) = 0 \\ \Rightarrow α + β = π / 2 + nπ, n \in I \\ \Rightarrow \sin (α + 2 β) = \sin (90^{\circ} + β) = \cos β (for n = 0) \end{array}$

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