Question

If tan (A + B) = $\sqrt{3}$ and tan (A − B) = $\frac{1}{\sqrt{3}}$, where A ≥ B and (A + B) is acute, then A = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer 1

(c) 45^o

$$\begin{gathered} \tan (A+\hspace{0.17em}B)=\sqrt{3}=\tan {60}^{\mathrm{o}}\left[\because \tan {60}^{\mathrm{o}}=\sqrt{3}\right] \\ \Rightarrow A+B={60}^{\mathrm{o}}...\left(1\right) \\ \tan (A-B)=\frac{1}{\sqrt{3}}=\tan {30}^{\mathrm{o}}\left[\because \tan {30}^{\mathrm{o}}=\frac{1}{\sqrt{3}}\right] \\ \Rightarrow A-B={30}^{\mathrm{o}}...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \mathrm{A}={45}^{\mathrm{o}}\mathrm{and}\mathrm{B}={15}^{\mathrm{o}} \\ \therefore A={45}^{\mathrm{o}} \end{gathered}$$