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

To do:

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

Solution:

We know that,

$\sin ^{2} A+\cos^2 A=1$.......(i)

$\operatorname{cosec} A=\frac{1}{\sin A}$......(ii)

Therefore,

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

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

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

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

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

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

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

Hence proved.    

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Updated on: 10-Oct-2022

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