Prove the following trigonometric identities:$ \frac{1+\sec \theta}{\sec \theta}=\frac{\sin ^{2} \theta}{1-\cos \theta} $

To do:

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

Solution:

We know that,

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

$\cos \theta=\frac{1}{\sec \theta}$.........(ii)

Therefore,

$\frac{1+\sec \theta}{\sec \theta}=\frac{1}{\sec \theta}+\frac{\sec \theta}{\sec \theta}$

$=\cos \theta+1$                [From (ii)]

Multiplying and dividing by ($1-\cos \theta$), we get,

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

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

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

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

Hence proved.       

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Updated on: 10-Oct-2022

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