Question

Prove that...
Please give the solution​

Answer 1

$\frac{ {cos}^{2} \theta }{1 - tan \theta} + \frac{ {sin}^{3} \theta }{sin \theta - cos \theta} \\ \\ = \frac{ {cos}^{3} \theta }{cos \theta - sin \theta} - \frac{ {sin}^{3} \theta }{cos \theta - sin \theta} \\ \\ = \frac{ {cos}^{3} \theta - {sin}^{3} \theta }{cos \theta - sin \theta} \\ \\ = \frac{(cos \theta - sin \theta)( {cos}^{2} \theta + cos \theta sin \theta + {sin}^{2} \theta) }{cos \theta - sin \theta} \\ \\ = 1 + sin \theta cos \theta$