Simplify: ^@ \dfrac{ \cos \theta } { 1 - \tan \theta } + \dfrac{ \sin^2 \theta } { \sin \theta - \cos \theta } ^@


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Answer:

^@ \sin \theta + \cos \theta ^@

Step by Step Explanation:
  1. ^@ \begin{align} & \dfrac{ \cos \theta } { 1 - \tan \theta } + \dfrac{ \sin^2 \theta } { \sin \theta - \cos \theta } \\ = & \dfrac{ \cos \theta \space \cos \theta } { \cos \theta - \cos \theta \tan \theta } + \dfrac{ \sin^2 \theta } { \sin \theta - \cos \theta } [\text{Multiply numerator and denominator of first term by } cos\theta]\\ = & \dfrac{ \cos^2 \theta } { \cos \theta - \sin \theta } + \dfrac{ \sin^2 \theta } { \sin \theta - \cos \theta } \space \left[\text{Since, } tan \theta = \dfrac{ sin \theta } { cos \theta } \right] \\ = & \dfrac{ \sin^2 \theta - \cos^2 \theta } { \sin \theta - \cos \theta } \\ = & \dfrac{ (\sin \theta - \cos \theta)(\sin \theta + \cos \theta) } { \sin \theta - \cos \theta } \space [\text{Since, } sin^2 \theta - cos^2 \theta = (\sin \theta - \cos \theta)(\sin \theta + \cos \theta) ] \\ = & \sin \theta + \cos \theta \end{align} ^@

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