Prove that:
$ \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta $

To do:

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

Solution:

We know that,

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

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

Therefore,

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

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

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

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

$=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta}=\frac{2}{\cos \theta}$

$=2 \sec \theta$

Hence proved.    

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Updated on: 10-Oct-2022

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