# Solve this

Question:

$\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 18^{\circ} \tan 35^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right)=?$

(a) $\frac{1}{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $\frac{2}{\sqrt{3}}$

Solution:

$\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 18^{\circ} \tan 35^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right)$

$=\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan \left(90^{\circ}-72^{\circ}\right) \tan \left(90^{\circ}-55^{\circ}\right) \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right)$

$=\left(\frac{\cos 38^{\circ} \operatorname{cosec}\left(90^{\circ}-38^{\circ}\right)}{\cot 72^{\circ} \cot 55^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right) \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\left(\frac{\cos 38^{\circ} \sec 38^{\circ}}{\cot 72^{\circ} \cot 55^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right) \quad\left(\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=\left(\frac{\cos 38^{\circ} \frac{1}{\cos 38^{\circ}}}{\frac{1}{\tan 72^{\circ}} \frac{1}{\tan 55^{\circ}} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}\right) \quad\left(\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{1}{\tan \theta}\right)$

$=\left(\frac{1}{\tan 60^{\circ}}\right)$

$=\frac{1}{\sqrt{3}} \quad\left(\because \tan 60^{\circ}=\sqrt{3}\right)$

Hence, the correct option is (b).