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

Answer 1

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

$[\frac{sinA}{sinB}=\frac{\sqrt{3}}{2}$ & $\frac{cosA}{cosB}=\frac{\sqrt{5}}{2}.$

$\Rightarrow sinA=\sqrt{3},sinB=2,cosA=\sqrt{5}$ & $cosB=2$

$tanA=\frac{sinA}{cosA}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{3}}{\sqrt{5}}$ & $tanB=\frac{sinB}{cosB}=\frac{2}{2}=1$

$tanA+tanB=\frac{\sqrt{3}}{\sqrt{5}}+1\Rightarrow \frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$

Option D is correct.