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Simplify: ^@1 - \dfrac { sin^3 \theta + \cos^3 \theta } { sin \theta + cos \theta }^@
Answer:
^@sin \theta \space cos \theta ^@
- ^@1 - \dfrac { sin^3 \theta + \cos^3 \theta } { sin \theta + cos \theta }^@ = ^@1 - \dfrac { (sin \theta + cos \theta) (sin^2 \theta + cos^2 \theta - sin \theta cos \theta) } { sin \theta + cos \theta }^@ ^@\space \space [\because x^3 + y^3 = (x + y)(x^2 + y^2 - xy)]^@
- Now, ^@sin \theta + cos \theta ^@ cancels out.
^@ \begin{align} &1 - (sin^2 \theta + cos^2 \theta - sin \theta cos \theta) \\ &\implies 1 - (1 - sin \theta cos \theta) \space \space [\because sin^2 \theta + cos^2 \theta = 1] \\ &\implies 1 - 1 + sin \theta cos \theta \\ &\implies sin \theta + cos \theta \end{align}^@
