Bangladesh
Simplify: ^@ \mathrm{cosec}^2 \theta \left( \dfrac{ \cos\theta - 1 } { 1 + \mathrm{cosec} \theta } \right) - \tan^2 \theta \left( \dfrac{ \mathrm{cosec} \theta - 1 } { 1 + \cos \theta } \right) ^@
Answer:
^@ 0 ^@
- On adding two fractions:
^@ \begin{align} & \mathrm{cosec}^2 \theta \left( \dfrac{ \cos\theta - 1 } { 1 + \mathrm{cosec} \theta } \right) - \tan^2 \theta \left( \dfrac{ \mathrm{cosec} \theta - 1 } { 1 + \cos \theta } \right) \\ = & \dfrac{ \mathrm{cosec}^2 \theta (\cos \theta - 1) (\cos \theta + 1 ) - \tan^2 \theta (\mathrm{cosec} \theta - 1) (\mathrm{cosec} \theta + 1 ) } { (1 + \mathrm{cosec} \theta ) (1 + \cos \theta )} \\ = & \dfrac{ \mathrm{cosec}^2 \theta(\cos^2 \theta - 1) - \tan^2\theta (\mathrm{cosec}^2 \theta - 1) } { (1 + \mathrm{cosec} \theta ) (1 + \cos \theta ) } \\ = & \dfrac{ \mathrm{cosec}^2 \theta \sin^2 \theta - \tan^2 \theta \cot^2 \theta } {(1 + \mathrm{cosec} \theta ) (1 + \cos \theta )} \\ = & \dfrac{ 1 - 1 } {(1 + \mathrm{cosec} \theta ) (1 + \cos \theta ) } \\ = & 0 \\ \end{align} ^@
