Question

Evaluate :
(ix) $\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}+\sqrt{3}\left\{\tan {10}^{\circ }\tan {40}^{\circ }\tan {30}^{\circ }\tan {50}^{\circ }\tan {80}^{\circ }\right\}$

Answer 1

We know,

$\cos (90-\theta )=\sin \theta$

$\tan (90-\theta )=\cot \theta$

$\cot \theta =\frac{1}{\tan \theta }$

$\cos {72}^{\circ }=\cos ({90}^{\circ }-{18}^{\circ })=\sin {18}^{\circ }$

$\tan {10}^{\circ }=\tan ({90}^{\circ }-{80}^{\circ })=\cot {80}^{\circ }$

$\tan {40}^{\circ }=\tan ({90}^{\circ }-{50}^{\circ })=\cot {50}^{\circ }$

$\csc {20}^{\circ }=\sec ({90}^{\circ }-{70}^{\circ })=\sec {70}^{\circ }$

To evaluate,

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

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

$=1+\frac{\sqrt{3}\left(\tan {30}^{\circ }\tan {50}^{\circ }\tan {80}^{\circ }\right)}{\left(\tan {80}^{\circ }\tan {50}^{\circ }\right)}$

$=1+\sqrt{3}\left(\tan {30}^{\circ }\right)$

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

$=1+1$

$=2$