Question

If $\frac{3 + cot 80^{o} cot ⁡ 20^{o}}{cot ⁡ 80^{o} + cot ⁡ 20^{o}}=$ $tan\left(\frac{a \pi }{b}\right),$ then the value of $a+b$ is

Answer 1

$\frac{3 \sin 80^{\circ} \sin 20^{\circ}+\cos 80^{\circ} \cos 20^{\circ}}{\sin 20^{\circ} \cos 80^{\circ}+\cos 20^{\circ} \sin 80^{\circ}}$
$=\frac{2 \sin 20^{\circ} \sin 80^{\circ}+\left(\cos 80^{\circ} \cos 20^{\circ}+\sin 80^{\circ} \sin 20\right)}{\sin \left(20^{\circ}+80^{\circ}\right)}$
$=\frac{\cos 60^{\circ}-\cos 100^{\circ}+\cos \left(80^{\circ}-20^{\circ}\right)}{\sin 100^{\circ}}$
$=\frac{2 \cos 60^{\circ}-\cos 100^{\circ}}{\sin 100^{\circ}}=\frac{1-\cos 100^{\circ}}{\sin 100^{\circ}}$
$=\frac{2 \sin { }^{2} 50^{\circ}}{2 \sin 50^{\circ} \cos 50^{\circ}=\tan 50^{\circ}}$
$=\tan \frac{5 \pi}{18}$
$\Rightarrow a+b=5+18=23$