Question

$\frac{2{\tan }^2{30}^{\circ }{\sec }^2{52}^{\circ }{\sin }^{2}38}{{\text{cosec}}^{2}70-{\tan }^{2}20}=$

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{2}{3}$

$\frac{2\times {\left(\frac{1}{\sqrt{3}}\right)}^2{\sec }^{2}52{\sin }^2(90-52)}{cose{c}^{2}70-{\tan }^2(90-70)}$, substitute the known values
$=\frac{2}{3}\times \frac{{\sec }^{2}52{\cos }^{2}52}{cose{c}^{2}70-{\cot }^{2}70}$
$=\frac{2}{3}\times \frac{1}{1}=\frac{2}{3}$

Note: $cose{c}^2\theta -{\cot }^2\theta =1$ and $\cos \theta \cdot \sec \theta =1$