Question

If $A={60}^o$ and $B={30}^o$, then prove that $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

If $A={60}^{\circ }$ and $B={30}^{\circ }$, then
$A-B={30}^{\circ }$
$\tan (A-B)=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
Now,
$\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\tan {60}^{\circ }-\tan {30}^{\circ }}{1+\tan {60}^{\circ }\tan {30}^{\circ }}$
$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}$
$=\frac{3-1}{\sqrt{3}}1+1$
$=1/\sqrt{3}$
$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$
Verified.