Verify the identity shown below.
$$\frac { 1 - \sin \theta } { 1 + \sin \theta } = 2 \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta - 1$$

Name: Verify the identity shown below. \[\frac { 1 - \sin \theta } { 1 + \sin \theta } = 2 \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta - 1\]
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\begin{aligned}\frac { 1 - \sin heta } { 1 + \sin heta } & = \frac { 1 - \sin heta } { 1 + \sin heta } \cdot \frac { 1 - \sin heta } { 1 - \sin heta } \& = \frac { ( 1 - \sin heta ) ^ { 2 } } { 1 - \sin ^ { 2 } heta } \& = \frac { 1 - 2 \sin heta + \sin ^ { 2 } heta } { \cos ^ { 2 } heta } \& = \frac { 1 } { \cos ^ { 2 } heta } - \frac { 2 \sin heta } { \cos ^ { 2 } heta } + \frac { \sin ^ { 2 } heta } { \cos ^ { 2 } heta } \& = \frac { 1 } { \cos ^ { 2 } heta } - \frac { 2 \sin heta } { \cos heta } \cdot \frac { 1 } { \cos heta } + \frac { \sin ^ { 2 } heta } { \cos ^ { 2 } heta } \& = \sec ^ { 2 } heta - 2 an heta \sec heta + an ^ { 2 } heta \& = \sec ^ { 2 } heta - 2 an heta \sec heta + \sec ^ { 2 } heta - 1 \& = 2 \sec ^ { 2 } heta - 2 an heta \sec heta - 1\end{aligned}

Verify the Identity Shown Below 1−sin⁡θ1+sin⁡θ=2sec⁡2θ−2sec⁡θtan⁡θ−1\frac { 1 - \sin \theta } { 1 + \sin \theta } = 2 \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta - 11+sinθ1−sinθ​=2sec2θ−2secθtanθ−1

Verify the Identity Shown Below $\frac { 1 - \sin \theta } { 1 + \sin \theta } = 2 \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta - 1$