Question

prove the following​

Answer 1

$\huge\underline\bold\blue{★Answer★}$

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