Question

If sin (α + β) = 1 and sin (α − β)$=\frac{1}{2}$, where 0 ≤ α, $\mathrm{\beta }\le \frac{\mathrm{\pi }}{2}$, then find the values of tan (α + 2β) and tan (2α + β).

Answer 1

$$\begin{gathered} \text{Given:} \\ \sin (\alpha +\beta )=1\mathrm{and}\sin (\alpha -\beta )=\frac{1}{2} \\ \Rightarrow \alpha +\beta =90°...\left(1\right) \\ \mathrm{and}\alpha -\beta =30°...\left(2\right) \\ \mathrm{By}\mathrm{adding}\mathrm{eq}\left(1\right)\text{and}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}: \\ 2\alpha =120° \\ \Rightarrow \alpha =60° \\ \mathrm{By}\mathrm{subtracting}\mathrm{eq}\left(2\right)\mathrm{from}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get}: \\ 2\beta =60° \\ \Rightarrow \beta =30° \\ \mathrm{Therefore}, \\ \tan (\alpha +2\beta )=\tan \left(60°+2\times 30°\right)=\tan 120°=-\sqrt{3} \\ \tan (2\alpha +\beta )=\tan \left(2\times 60°+30°\right)=\tan 150°=-\frac{1}{\sqrt{3}} \end{gathered}$$