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

2sec2θ 1 0 2tan2θ

2sec2θ

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