Question

Problem in solving. I am facing trouble in solving the second part.

Answer 1

$\frac{cosec ^{2}(90- \alpha)-tan^{2}\alpha}{2(cos^{2}48+cos^{2}42)} -  \frac{2.tan^{2}30.sec^{2}52.sin^{2}38}{cosec^{2}70-tan^{2}20}$
$=\frac{sec ^{2}\alpha-tan^{2}\alpha}{2(sin^{2}42+cos^{2}42)} - \frac{2.(1/ \sqrt{3})^{2}.cosec^{2}38.sin^{2}38}{sec^{2}20-tan^{2}20}$
$=\frac{1}{2(1)} - \frac{2.(1/3).1}{1} = \frac{1}{2}-\frac{2}{3} = -\frac{1}{6}$