Question

If $\sin (A+B)=1$ and $\sin (A-B)$ = $\frac{1}{\sqrt{3}}$, find the value of $\tan A+\tan B$.

Answer 1

$\sin (A+B)=1⟹A+B=\pi /2⟹A=\pi /2-B$

$\sin (A-B)=\frac{1}{\sqrt{3}}⟹\sin (\pi /2-2B)=\frac{1}{\sqrt{3}}⟹\cos 2B=\frac{1}{\sqrt{3}}⟹\sin 2B=\sqrt{\frac{2}{3}}$

$\sin 2B=\sqrt{\frac{2}{3}}⟹\frac{2\tan B}{1+{\tan }^{2}B}=\sqrt{\frac{2}{3}}$

$⟹\frac{1+{\tan }^{2}B}{\tan B}=\sqrt{6}⟹\cot B+\tan B=\sqrt{6}⟹\cot (\pi /2-A)+\tan B=\sqrt{6}$

$⟹\tan A+\tan B=\sqrt{6}$