Question

If $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}$and $\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2},0<A,B<\frac{\pi }{2},$ then find the value of $\tan A+\tan B$

• A. $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}$
• B. $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}}$
• C. $\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$
• D. $\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$

Given $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}$and $\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}$
$\Rightarrow \sin A=\frac{\sqrt{3}}{2}\sin B$ and $\cos A=\frac{\sqrt{5}}{2}\cos B$
Squaring and adding both equations,
${\sin }^{2}A+{\cos }^{2}A=\frac{3}{4}{\sin }^{2}B+\frac{5}{4}{\sin }^{2}B$
$\Rightarrow 3{\sin }^{2}B+5{\cos }^{2}B=4\Rightarrow {\cos }^{2}B=\frac{1}{2}\Rightarrow {\tan }^{2}B=1\Rightarrow \tan B=1$
Now using first equation, $\sin A=\frac{\sqrt{3}}{2\sqrt{2}}\Rightarrow \tan A=\frac{\sqrt{3}}{\sqrt{5}}$
Hence $\tan A+\tan B=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$