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Compound angles

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Question

If $\sin (α + β) = 1, \sin (α - β) = 1 / 2; α, β \in [π / 2]$ then $\tan (α + 2 β) \tan (2 α + β)$ is equal to

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Solution

We have,
$\begin{array}{l} \sin (α + β) = 1 and, \sin (α - β) = \frac{1}{2} \\ \Rightarrow α + β = \frac{π}{2} and, α - β = \frac{π}{6} \\ \Rightarrow α = \frac{π}{3}, β = \frac{π}{6} \\ ∴ \tan (α + 2 β) \tan (2 α + β) \\ = \tan (\frac{2 π}{3}) \tan \frac{5 π}{6} = (- \cot \frac{π}{6}) (- \cot \frac{π}{3}) = 1 \end{array}$

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