Solve the equation for exact solutions.
-$$( \sec \alpha + \tan \alpha ) ^ { 2 } = \frac { 1 + \sin \alpha } { 1 - \sin \alpha }$$

Name: Solve the equation for exact solutions. -\[( \sec \alpha + \tan \alpha ) ^ { 2 } = \frac { 1 + \sin \alpha } { 1 - \sin \alpha }\]
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\begin{array} { l } ( \sec \alpha + an \alpha ) ^ { 2 } = \sec ^ { 2 } \alpha + 2 \sec \alpha an \alpha + an ^ { 2 } \alpha = \frac { 1 } { \cos ^ { 2 } \alpha } + \frac { 2 \sin \alpha } { \cos ^ { 2 } \alpha } + \frac { \sin ^ { 2 } \alpha } { \cos ^ { 2 } \alpha } = \\frac { 1 + 2 \sin \alpha + \sin ^ { 2 } \alpha } { \cos ^ { 2 } \alpha } = \frac { ( 1 + \sin \alpha ) ^ { 2 } } { 1 - \sin ^ { 2 } \alpha } = \frac { ( 1 + \sin \alpha ) ^ { 2 } } { ( 1 - \sin \alpha ) ( 1 + \sin \alpha ) } = \frac { 1 + \sin \alpha } { 1 - \sin \alpha }\end{array}

Solve the Equation for Exact Solutions (sec⁡α+tan⁡α)2=1+sin⁡α1−sin⁡α( \sec \alpha + \tan \alpha ) ^ { 2 } = \frac { 1 + \sin \alpha } { 1 - \sin \alpha }(secα+tanα)2=1−sinα1+sinα​

Solve the Equation for Exact Solutions $( \sec \alpha + \tan \alpha ) ^ { 2 } = \frac { 1 + \sin \alpha } { 1 - \sin \alpha }$