Question

prove that (1 +sin theta - cos theta /1+sin theta +cos theta) ^2 =1-cos/1+cos​

Answer 1

$\left(\dfrac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta}\right)^2 =\\ \\ \\ = \dfrac{1^2+\sin^2 \theta+\cos^2 \theta+2\sin \theta-2\cos \theta-2\sin \theta\cos \theta}{1^2+\sin^2 \theta+\cos^2 \theta+2\sin \theta+2\cos \theta+2\sin \theta\cos \theta }=\\ \\ \\ =\dfrac{2(1+\sin \theta-\cos\theta-\sin \theta\cos \theta)}{2(1+\sin \theta+\cos \theta+\sin \theta\cos\theta)} =\\ \\ \\ = \dfrac{(\sin \theta)(1-\cos \theta)+1-\cos\theta}{(\sin \theta)(1+\cos \theta)+1+\cos\theta} = \\ \\\\ =\dfrac{(1-\cos \theta)(\sin \theta+1)}{(1+\cos \theta)(\sin \theta+1)} =\\ \\ \\= \dfrac{1-\cos \theta}{1+\cos \theta}$