Question

If $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}$ and $\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2},0<A,B<\pi /2,$ then $\tan A+\tan B$ is equal to

• A. $d\frac{\sqrt{3}}{\sqrt{5}}$
• B. $\frac{\sqrt{5}}{\sqrt{3}}$
• C. $1$
• D. $\frac{(\sqrt{5}+\sqrt{3})}{\sqrt{5}}$

Answer 1

The correct option is D $\frac{(\sqrt{5}+\sqrt{3})}{\sqrt{5}}$

$\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}\Rightarrow \frac{\sin {}^{2}A}{\sin {}^{2}B}=\frac{3}{4}\Rightarrow \sin {}^{2}A=3\frac{\sin {}^{2}B}{4}-\left(i\right)\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}\frac{\cos {}^{2}A}{\cos {}^{2}B}=\frac{5}{4}\Rightarrow \cos {}^{2}A=5\frac{\cos {}^{2}B}{4}-\left(ii\right)\left(i\right)+\left(ii\right)=1\therefore 3\frac{\sin {}^{2}B}{4}+5\frac{\cos {}^{2}B}{4}=1\therefore 3\sin {}^{2}B+5(1-\sin {}^{2}B)=4\therefore 1=2\sin {}^{2}B\Rightarrow \sin B=\frac{1}{\sqrt{2}}B=45\sin A=\frac{\sqrt{6}}{2}\cos B=\cos 45=\frac{1}{\sqrt{2}}\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}\cos A=\frac{\sqrt{2}\sqrt{5}}{2}\Rightarrow \tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=\frac{\frac{\sqrt{6}}{2}}{\sqrt{2}\frac{\sqrt{5}}{2}}+\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3}}{\sqrt{5}}+1=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$