Prove that:
$ \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=2 \operatorname{cosec} \theta $

To do:

We have to prove that \( \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=2 \operatorname{cosec} \theta \).

Solution:

We know that,

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

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

Therefore,

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

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

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

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

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

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

$=\frac{2}{\sin \theta}$

$=2 \operatorname{cosec} \theta$

Hence proved.  

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Updated on: 10-Oct-2022

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