Question

$\frac{2{\tan }^{2}30°{\sec }^{2}52°{\sin }^{2}38°}{({\mathrm{cosec}}^{2}70°-{\tan }^{2}20°)}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) $\frac{1}{2}$

Answer 1

(b) $\frac{2}{3}$

$$\begin{gathered} \mathrm{W}\text{e have:} \\ \left[\frac{2{\tan }^2{30}^0{\sec }^2{52}^0{\sin }^2{38}^0}{\cos e{c}^2{70}^0-{\tan }^2{20}^0}\right] \\ =\left[\frac{2\times {\left(\frac{1}{\sqrt{3}}\right)}^2{\sec }^2{52}^0\left\{{\sin }^2({90}^0-{52}^0)\right\}}{\left\{\cos {\text{ec}}^2({90}^0-{20}^0)\right\}-{\tan }^2{20}^0}\right] \\ =[\frac{2}{3}\times \frac{{\sec }^2{52}^0.{\cos }^2{52}^0}{{\sec }^2{20}^0-{\tan }^2{20}^0}][\because \sin ({90}^0-\theta )=\cos \theta \text{and}\cos \text{ec}({90}^0-\theta )=\sec \theta ] \\ =\frac{2}{3}\times \frac{1}{1}[\because {\sec }^2\theta -{\tan }^2\theta =1] \\ =\frac{2}{3} \end{gathered}$$