Question

$\frac{{\sin }^{}{18}^o}{\cos {72}^o}+\sqrt{3}\left(\tan {10}^o.\tan {30}^o.\tan {40}^o.\tan {50}^o.\tan {80}^o.\right)$

Answer 1

$\frac{sin{18}^o}{cos{72}^o}+\sqrt{3}(tan{10}^o.tan{30}^o.tan{40}^o.tan{50}^o.tan{80}^o)$

$=\frac{sin{18}^o}{\sin \left({90}^{\circ }-{72}^o\right)}+\sqrt{3}\left(cot\right({90}^{\circ }-{10}^o).tan{30}^o.cot({90}^{\circ }-{40}^o).tan{50}^o.tan{80}^o)$

$=\frac{sin{18}^o}{\sin {18}^{\circ }}+\sqrt{3}(cot{80}^{\circ }.tan{80}^o.cot{50}^{\circ }.tan{50}^o.tan{30}^o)$

$1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)=2$