Question

Prove that sec2θ − cos2θ = tan2θ + sin2θ​

Answer 1

Answer:

get answer by solving rhs

Step-by-step explanation:

$tan^{2} \alpha + sin^{2} \alpha$

=$(sec^{2}\alpha -1)+(1 - cos^{2})$       ( ∵ $tan^{2}\alpha = sec^{2}\alpha - 1 , sin^{2}\alpha = 1 - cos^{2}\alpha$ )

= sec2O -1 +1 -cos2O

= sec2O - cos2O

= l.h.s.

Answer 2

The explanation is given above

I hope u understand

Thank u