Question

$\frac{\cos 38°\mathrm{cosec}52°}{\tan 18°\tan 35°\tan 60°\tan 72°\tan 55°}=?$
(a) $\sqrt{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}}$

Answer 1

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

$$\begin{gathered} \text{We have:} \\ \left[\frac{\cos {38}^0\cos \text{ec}{52}^0}{\tan {18}^0\tan {35}^0\tan {60}^0\tan {72}^0\tan {55}^0}\right] \\ =\left[\frac{\cos {38}^0\cos \text{ec}({90}^0-{38}^0)}{\tan {18}^0\tan {35}^0\times \sqrt{3}\times \tan ({90}^0-{18}^0)\tan ({90}^0-{35}^0)}\right]\left[\begin{array}{l}\because \cos \text{ec}({90}^0-\theta )=\sec \theta \text{and}\\ \tan ({90}^0-\theta )=\cot \theta \end{array}\right] \\ =\left[\frac{\cos {38}^0\sec {38}^0}{\tan {18}^0\tan {35}^0\times \sqrt{3}\times \cot {18}^0\cot {35}^0}\right] \\ =\left[\frac{\frac{1}{\sec {38}^0}\times \sec {38}^0}{\frac{1}{\cot {18}^0}\frac{1}{\cot {35}^0}\times \sqrt{3}\cot {18}^0\cot {35}^0}\right] \\ =\frac{1}{\sqrt{3}} \end{gathered}$$