Prove that:$ \frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta} $

To do:

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

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,

$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{\{(1+\cos \theta)+\sin \theta\}\{(1+\cos \theta)+\sin \theta\}}{\{(1+\cos \theta)-\sin \theta\}\{(1+\cos \theta)+\sin \theta\}}$    (Multiplying numerator and denominator by $1+\cos \theta+\sin \theta$)

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

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

Hence proved.