Question

If $sinA=\frac{\sqrt{3}}{2}$ and $cosB=\frac{\sqrt{3}}{2}$, then the value of : $\frac{tanA-tanB}{1+tanAtanB}$ is $\frac{1}{\sqrt{m}}=\frac{\sqrt{3}}{3}$
value of m is

Answer 1

$\sin A=\frac{\sqrt{3}}{2}$
$\sin A=\sin 60$
$A={60}^{\circ }$
$\cos B=\frac{\sqrt{3}}{2}$
$\cos B=\sin 30$
$B={30}^{\circ }$
Now, $\frac{tanA-tanB}{1+tanAtanB}$
$\frac{tan60-tan30}{1+tan60tan30}$
$\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)}$
$\frac{3-1}{2\sqrt{3}}$
$\frac{1}{\sqrt{3}}$
$\frac{\sqrt{3}}{3}$