Question

If $\sin A=\frac{\sqrt{3}}{2}$ and $\cos B=\frac{\sqrt{3}}{2}$, find the value of $\frac{\tan A-\tan B}{1+\tan A\tan B}$. is $\frac{1}{\sqrt{m}}$
, m is

Answer 1

Given $\sin A=\frac{\sqrt{3}}{2}$

$\sin A=\sin 60$

$A={60}^{\circ }$

$\cos B=\frac{\sqrt{3}}{2}$

$\cos B=\cos 30$

$B={30}^{\circ }$

Now, $\frac{\tan A-\tan B}{1+\tan A\tan B}$

= $\frac{\tan 60-\tan 30}{1+\tan 60\tan 30}$

= $\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)}$

= $\frac{\frac{2}{\sqrt{3}}}{2}$

= $\frac{1}{\sqrt{3}}$