Question

Prove that ${\left(\frac{1+sin\Theta -cos\Theta }{1+sin\Theta +cos\Theta }\right)}^2=\frac{1-cos\Theta }{1+cos\Theta }$

Answer 1

consider the problem

$L.H.S$ $={\left[\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }\right]}^2$

$\begin{array}{c}={\left[\frac{\left(1+\sin \theta \right)-\cos \theta }{\left(1+\sin \theta \right)+\cos \theta }\times d\frac{\left(1+\sin \theta \right)-\cos \theta }{\left(1+\sin \theta \right)-\cos \theta }\right]}^2\\ ={\left[\frac{1+{\sin }^2\theta +{\cos }^2\theta +2\sin \theta -2\sin \theta .\cos \theta -2\cos \theta }{1+{\sin }^2\theta +2\sin \theta -{\cos }^2\theta }\right]}^2\\ ={\left[\frac{2+2\sin \theta -2\sin \theta .\cos \theta -2\cos \theta }{2{\sin }^2\theta +2\sin \theta }\right]}^2\\ ={\left[\frac{2\left(1+\sin \theta \right)-2\cos \theta \left(\sin \theta +1\right)}{2\sin \theta \left(\sin \theta +1\right)}\right]}^2\\ ={\left[\frac{\left(1+\sin \theta \right)\left(1-\cos \theta \right)}{\sin \theta \left(\sin \theta +1\right)}\right]}^2\\ ={\left(d\frac{1-\cos \theta }{\sin \theta }\right)}^2\\ =\frac{{\left(1-\cos \theta \right)}^2}{{\sin }^2\theta }\\ =\frac{{\left(1-\cos \theta \right)}^2}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)}\end{array}$

$L.H.S=R.H.S$