Question

$\frac{\sin (A+B)}{\cos A\cos B}=\tan A+\tan B$

Answer 1

Given: $A={60}^{\circ },B={30}^{\circ }$
Now, $\frac{\sin (A+B)}{\cos A\cos B}=\tan A+\tan B$
$L.H.S.=\frac{\sin (A+B)}{\cos A\cos B}$
= $\frac{\sin ({60}^0+{30}^0)}{\cos {60}^0\cos {30}^0}$
= $\frac{\sin {90}^0}{\cos {60}^0\cos {30}^0}$
= $\frac{1}{\cos {60}^0\cos {30}^0}$
= $\sec {60}^0\sec {30}^0$
= $\frac{2}{\sqrt{3}}.2$
= $\frac{4}{\sqrt{3}}$
$R.H.S=\tan A+\tan B$
= $\tan {60}^0+\tan {30}^0$
= $\sqrt{3}+\frac{1}{\sqrt{3}}$
= $\frac{3+1}{\sqrt{3}}$
= $\frac{4}{\sqrt{3}}$
$\therefore L.H.S=R.H.S$

$\therefore \frac{\sin (A+B)}{\cos A\cos B}=\tan A+\tan B$