Question

Given
$\frac{\sin A}{sinB}=\frac{\sqrt{3}}{2},\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2},0<A,B<\frac{\pi }{2}$, then

• A. $\tan A=\sqrt{\frac{3}{5}}$
• B. $\tan A=\sqrt{\frac{5}{3}}$
• C. $\tan A=2$
• D. $\tan B=2$

Answer 1

The correct option is A $\tan A=\sqrt{\frac{3}{5}}$

straight away we get $\sin B=\frac{2sinA}{\sqrt{3}}and\cos B=\frac{2cosA}{\sqrt{5}}$

now ${\sin }^{2}B+{\cos }^{2}B=1$

so we get $1=4(\frac{{\sin }^{2}A}{3}+\frac{{\cos }^{2}A}{5})$

we divide throughout by ${\cos }^{2}A$ and put ${\sec }^{2}A=1+{\tan }^{2}A$

$\frac{{\sec }^{2}A}{4}=(\frac{{\tan }^{2}A}{3}+\frac{1}{5})$

$\frac{{\tan }^{2}A}{3}=\frac{1+{\tan }^{2}A}{4}-\frac{1}{5}$
$(\frac{1}{3}-\frac{1}{4}){\tan }^{2}A=\frac{1}{4}-\frac{1}{5}$
$\frac{{\tan }^{2}A}{12}=\frac{1}{20}$

$\tan A=\sqrt{\frac{3}{5}}$