Solve this

Question:

$\frac{1-\sin \theta}{1+\sin \theta}=(\sec \theta-\tan \theta)^{2}$

 

Solution:

$\frac{1-\sin \theta}{1+\sin \theta}$

$=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$

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

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

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

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

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

$=(\sec \theta-\tan \theta)^{2}$

 

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