Prove this identity.
\(\frac{1+sin\theta}{1-sin\theta}=(sec\theta+tan\theta)^2\)
From everything I've tried, I don't think this is an identity but maybe I'm wrong.
\(\dfrac{1+\sin(\theta)}{1-\sin(\theta)} \cdot \dfrac{1+\sin(\theta)}{1+\sin(\theta)} = \\ \dfrac{(1+\sin(\theta))^2}{1-\sin^2(\theta)} = \\ \left(\dfrac{1+\sin(\theta)}{\cos(\theta)}\right)^2 = \\ (\sec(\theta)+\tan(\theta))^2\)
(sec x + tanx ) ^2 =
(1/ cosx + sinx / cosx)^2 =
(1 + sin x)^2 (1+ sin x) (1 + sin x) (1 + sin x) (1 + sinx) 1 + sin x
_________ = _________________ = ___________________ = ___________
cos^2x 1 - sin^2x (1 + sin x)(1 - sin x) 1 - sin x
+895 
