Bahrain
Simplify: ^@ \dfrac{ \cos^2 \theta } { 1 - \tan \theta } + \dfrac{ \sin^3 \theta } { \sin \theta - \cos \theta } ^@
Answer:
^@ 1 + sin \theta \space cos \theta ^@
- ^@ \begin{align} & \dfrac{ \cos^2 \theta } { 1 - \tan \theta } + \dfrac{ \sin^3 \theta } { \sin \theta - \cos \theta } \\ = & \dfrac{ \cos \theta \cos^2 \theta } { \cos \theta - \cos \theta \tan \theta } + \dfrac{ \sin^3 \theta } { \sin \theta - \cos \theta } [\text{Multiply numerator and denominator of first term by } cos\theta]\\ = & \dfrac{ \cos^3 \theta } { \cos \theta - \sin \theta } - \dfrac{ \sin^3 \theta } { \cos\theta - \sin\theta } \\ = & \dfrac{ \cos^3 \theta - \sin^3 \theta } { \cos \theta - \sin \theta } \\ = & \dfrac{ (\cos \theta - \sin \theta) (\cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta) } { \cos \theta - \sin \theta } \space [\text{Since, } x^3 - y^3 = (x + y)(x^2 + y^2 +xy)] \\ = & 1 + sin \theta \space cos \theta \space [\text{Since, } sin^2 \theta + cos^2 \theta = 1] \end{align} ^@
