Question

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

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

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

(c) $\sqrt{3}$

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

Answer 1

$$\begin{gathered} \left(\frac{\cos 38°\mathrm{cosec}52°}{\tan 18°\tan 35°\tan 60°\tan 72°\tan 55°}\right) \\ =\left(\frac{\cos 38°\mathrm{cosec}52°}{\tan \left(90°-72°\right)\tan \left(90°-55°\right)\tan 60°\tan 72°\tan 55°}\right) \\ =\left(\frac{\cos 38°\mathrm{cosec}\left(90°-38°\right)}{\cot 72°\cot 55°\tan 60°\tan 72°\tan 55°}\right)\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =\left(\frac{\cos 38°\sec 38°}{\cot 72°\cot 55°\tan 60°\tan 72°\tan 55°}\right)\left(\because \mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =\left(\frac{\cos 38°{\displaystyle \frac{1}{\cos 38°}}}{{\displaystyle \frac{1}{\tan 72°}}{\displaystyle \frac{1}{\tan 55°}}\tan 60°\tan 72°\tan 55°}\right)\left(\because \sec \theta =\frac{1}{\cos \theta }\mathrm{and}\cot \theta =\frac{1}{\tan \theta }\right) \\ =\left(\frac{1}{{\displaystyle \tan 60°}}\right) \\ =\frac{1}{\sqrt{3}}\left(\because \tan 60°=\sqrt{3}\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{b}\right). \end{gathered}$$