Question

Given $A={60}^0$ and $B={30}^0$, then
$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

If true enter 1 else 0

Answer 1

Given, $A={60}^{\circ },B={30}^{\circ }$
Now, $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$
L.H.S: $\tan (A-B)=\tan (60-30)=\tan 30=\frac{1}{\sqrt{3}}$
R.H.S: $\frac{\tan A-\tan B}{1+\tan A\tan B}$
$=\frac{\tan 60-\tan 30}{1+\tan 60\tan 30}$
$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}$
$=\frac{3-1}{2\sqrt{3}}$
$=\frac{1}{\sqrt{3}}$
Thus, $L.H.S=R.H.S$