Question

$\left(\frac{2{\tan }^{2}30°{\sec }^{2}52°{\sin }^{2}38°}{{\mathrm{cosec}}^{2}70°-{\tan }^{2}20°}\right)=?$

(a) 2

(b) 1

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

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

Answer 1

$$\begin{gathered} \left(\frac{2{\tan }^{2}30°{\sec }^{2}52°{\sin }^{2}38°}{{\mathrm{cosec}}^{2}70°-{\tan }^{2}20°}\right) \\ =\left(\frac{2{\tan }^{2}30°{\left(\sec \left(90°-38°\right)\right)}^2{\sin }^{2}38°}{{\left(\mathrm{cosec}\left(90°-20°\right)\right)}^2-{\tan }^{2}20°}\right) \\ =\left(\frac{2{\tan }^{2}30°{\mathrm{cosec}}^{2}38°{\sin }^{2}38°}{{\sec }^{2}20°-{\tan }^{2}20°}\right)\left(\because \sec \left(90°-\theta \right)=\mathrm{cosec}\theta \mathrm{and}\mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =\left(\frac{2{\tan }^{2}30°{\mathrm{cosec}}^{2}38°{\sin }^{2}38°}{1}\right)\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sec }^2\theta -{\tan }^2\theta =1\right) \\ =2{\tan }^{2}30°\frac{1}{{\sin }^{2}38°}{\sin }^{2}38°\left(\because \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right) \\ =2{\tan }^{2}30° \\ =2{\left(\frac{1}{\sqrt{3}}\right)}^2\left(\because \tan 30°=\frac{1}{\sqrt{3}}\right) \\ =\frac{2}{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$