Question

If $\frac{3+cot80^{o}cot20^o}{cot80^o+cot20^o}=$ $tan\left(\frac{a\pi }{b}\right),$ then the value of $a+b$ is

Answer 1

Answer: 23

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