Question

Question number 12. ​

Answer 1

Hii its tom85

{1/(sec^2x - cos^2x)}=cos^2x/(1-cos^4x)_____(1

{1/(cosec^2x - sin^2x)}=sin^2x/(1-sin^4x)___(2

now

LHS

putting the value from above two equation we get

{(cos^2x/(1-cos^4x )+sin^2x/(1-sin^4x)}sin^2xcos^2x

{cos^2x/(sin^2x(1+cos^2x)+sin^2x/cos^2x(1+sin^2x)}sin^2xcos^2x

{cot^2x/(1+sin^2x)+tan^2x/1+cos^2x}sin^2xcos^2x

cos^4x/1+sin^2x + sin^4x/1+cos^2x

(cos^4x+cos^6x+sin^4x+sin^6x)/

(2+sin^2xcos^2x)

(1+2sin^2xcos^2x+1-3sin^2xcos^2x)/(2+sin^2xcos^2x)

(2-sin^2xcos^2x)/(2+sin^2xcos^2x)

RHS

__________/\__________☺️

hope it helps you dude