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

To do:

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

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

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

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

$=\sin \theta \cos \theta\left[\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\right]$

$=\sin \theta \cos \theta\left[\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}\right]$

$=\sin \theta \cos \theta\left[\frac{2}{1-\sin ^{2} \theta}\right]$

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

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

$=2 \tan \theta$

Hence proved.     

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Updated on: 10-Oct-2022

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