Question

cosec ^2(90-theta)- tan^2theta

Answer 1

cosec square (90-theta)=sec square theta then
sec square theta- tan square theta =0

Answer 2

Answer:

Given Expression:

$\frac{cosec^2(90^{\circ}-\theta)-tan^2\,\theta}{2(cos^2\,48^{\circ}+cos^2\,42^{\circ})}-\frac{2tan^2\,30^{\circ}\:sec^2\,52^{\circ}\:sin^2\,38^{\circ}}{cosec^2\,70^{\circ}-tan^2\,20^{\circ}}$

$\implies\frac{(cosec(90^{\circ}-\theta))^2-tan^2\,\theta}{2((cos\,48^{\circ}))^2+cos^2\,42^{\circ})}-\frac{2tan^2\,30^{\circ}\:(sec\,52^{\circ})^2\:sin^2\,38^{\circ}}{(cosec\,70^{\circ})^2-tan^2\,20^{\circ}}$

$\implies\frac{(sec\,\theta)^2-tan^2\,\theta}{2((cos\,(90^{\circ}-42^{\circ}))^2+cos^2\,42^{\circ})}-\frac{2tan^2\,30^{\circ}\:(sec\,(90^{\circ}-38^{\circ}))^2\:sin^2\,38^{\circ}}{(cosec\,(90^{\circ}-20^{\circ}))^2-tan^2\,20^{\circ}}$

$\implies\frac{sec^2\,\theta-tan^2\,\theta}{2(sin^2\,42^{\circ}+cos^2\,42^{\circ})}-\frac{2tan^2\,30^{\circ}\:(sec\,(90^{\circ}-38^{\circ}))^2\:sin^2\,38^{\circ}}{sec^2\,20^{\circ}-tan^2\,20^{\circ}}$

$\implies\frac{1}{2(1)}-\frac{2\times(\frac{1}{\sqrt{3}})^2\:cosec^2\,38^{\circ}\:sin^2\,38^{\circ}}{1}$

$\implies\frac{1}{2}-\frac{2}{3}$

$\implies\frac{3-4}{6}$

$\implies\frac{-1}{6}$