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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.