Prove the following identities:$ \left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\frac{1-\cos \theta}{1+\cos \theta} $

To do:

We have to prove that \( \left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\frac{1-\cos \theta}{1+\cos \theta} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

$=\frac{2+2 \sin \theta-2 \cos \theta-2 \sin \theta \cos \theta}{2+2 \sin \theta+2 \cos \theta+2 \sin \theta \cos \theta}$

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

$=\frac{1(1+\sin \theta)-\cos \theta(1+\sin \theta)}{1(1+\sin \theta)+\cos \theta(1+\sin \theta)}$

$=\frac{(1+\sin \theta)(1-\cos \theta)}{(1+\sin \theta)(1+\cos \theta)}$

$=\frac{1-\cos \theta}{1+\cos \theta}$

 Hence proved.