Thailand
Simplify ^@ \dfrac{ ( 1 + \cot \theta + \tan \theta) (\sin \theta - \cos \theta) } { \sec^3 \theta - \mathrm{cosec}^3 \theta } ^@
Answer:
^@ sin^2\theta \space cos^2\theta ^@
- ^@ \begin{align} & \dfrac{ ( 1 + \cot \theta + \tan \theta) (\sin \theta - \cos \theta) } { \sec^3 \theta - \mathrm{cosec}^3 \theta } \\ \\ = & \dfrac{ \left( 1 + \dfrac{ cos\theta } { sin\theta } + \dfrac{ sin\theta } { cos\theta } \right) \left( sin\theta - cos\theta \right) } { (sec\theta - cosec\theta)(sec^2\theta + sec\theta \times cosec\theta + cosec^2\theta) } \space [\text{Since, } a^3 - b^3 = (a - b)(a^2 + ab + b^2] \\ \\ = & \dfrac{ \left( \dfrac{ sin\theta \space cos\theta + cos^2\theta + sin^2\theta } { sin\theta \space cos\theta } \right) \left( sin\theta - cos\theta \right) } { \left( \dfrac { 1 } { cos\theta } - \dfrac{ 1 } { sin\theta } \right) \left( \dfrac{ 1 } { cos^2\theta } + \dfrac{ 1 } { sin\theta \space cos\theta } + \dfrac{ 1 } { cos^2\theta } \right) } \space \left[ \text{Since, } sec\theta = \dfrac{ 1 } { cos\theta } \text{ and } cosec\theta = \dfrac{ 1 } { sin\theta } \right] \\ \\ \\ = & \dfrac{ \left( \dfrac{ 1 + sin\theta \space cos\theta } { sin\theta \space cos\theta } \right) \left( sin\theta - cos\theta \right) } { \dfrac{ ( sin\theta - cos\theta )(sin^2\theta + sin\theta \space cos\theta + cos^2\theta) } { (sin\theta \space cos\theta) (sin^2\theta \space cos^2\theta) } } \\ \\ = & \dfrac{ (1 + sin\theta \space cos\theta)(sin\theta - cos\theta) (sin^3\theta \space cos^3\theta) } { (sin\theta \space cos\theta)(sin\theta - cos\theta)(1 + sin\theta \space cos\theta) } \\ \\ = & sin^2\theta \space cos^2\theta \end{align} ^@
