Question

Prove that 1 - sec θ + tan^2 θ / sin θ ( sec θ - 1 ) = 1 / sin θ - cos θ​

Answer 1

$\frac{1 - \sec \theta + { \tan}^{2} \theta }{ \sin \theta( \sec \theta - 1)} \\ \\ = \frac{ { \sec}^{2} \theta - \sec \theta }{\sin \theta( \sec \theta - 1)} \\ \\ = \frac{ \sec \theta}{ \sin \theta} \\ \\ = \frac{1}{sin \theta \: cos \theta}$