1/(1+cos(90^o-θ ))+1/(1-cos(90^o - θ ))=_________.

$\frac{1}{1+cos(90^o-\theta )}+\frac{1}{1-cos(90^o - \theta )}$

We know , cos (90º - θ ) = sin θ

= \(\frac{1}{1 + sin θ}\) + \(\frac{1}{1 - sin θ}\)

= \(\frac{2}{1 - sin^2θ}\)

= \(\frac{2}{cos^2θ}\)

= 2 sec² θ