Prove the following trigonometric identities:$ \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\operatorname{cosec} \theta-\cot \theta $

To do:

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

Solution:

We know that,

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

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

$\cot A=\frac{\cos A}{\sin A}$........(iii)

Therefore,

$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}\times \frac{1-\cos \theta}{1-\cos \theta}}$     (Multiply and divide by $1-\cos \theta$)

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

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

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

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

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

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

$=\operatorname{cosec} \theta-\cot \theta$    (From (ii) and (iii))

Hence proved.  

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Updated on: 10-Oct-2022

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