Question:
$\frac{1-\sin \theta}{1+\sin \theta}=(\sec \theta-\tan \theta)^{2}$
Solution:
$\frac{1-\sin \theta}{1+\sin \theta}$
$=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta} \quad\left(\sin ^{2} \theta+\cos ^{2} \theta=1\right)$
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$
$=\left(\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\right)^{2}$
$=(\sec \theta-\tan \theta)^{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.