Vietnam
Simplify: ^@ \cot^2 \theta \left( \dfrac{ \sec \theta - 1 } { 1 + \sin \theta } \right) + \sec^2 \theta \left( \dfrac{ \sin\theta - 1 } { 1 + \sec \theta } \right) ^@
Answer:
^@ 0 ^@
- On adding the two fractions, we have:
^@ \begin{align} & \cot^2 \theta \left( \dfrac{ \sec \theta - 1 } { 1 + \sin \theta } \right) + \sec^2 \theta \left( \dfrac{ \sin\theta - 1 } { 1 + \sec \theta } \right) \\ = & \dfrac{ \cot^2 \theta (\sec \theta - 1) (\sec \theta +1 ) + \sec^2 \theta (\sin \theta - 1) (\sin \theta +1 ) } { (1 + \sin \theta ) (1 + \sec \theta ) } \\ = & \dfrac{ \cot^2 \theta (\sec^2 \theta - 1) + \sec^2 \theta (\sin^2 \theta - 1) } { (1 + \sin \theta ) (1 + \sec \theta ) } \\ = & \dfrac{ \cot^2 \theta \tan^2 \theta - \sec^2 \theta \cos^2 \theta } { (1 + \sin \theta ) (1 + \sec \theta ) } \\ = & \dfrac{ 1 - 1 } { (1 + \sin \theta ) (1 + \sec \theta ) } \\ = & 0 \\ \end{align} ^@
