Prove the following trigonometric identities:$ \frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta} $

To do:

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

Solution:

We know that,

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

Therefore,

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

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

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

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

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

Hence proved.  

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Updated on: 10-Oct-2022

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