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

To do:

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

Solution:

We know that,

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

Therefore,

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

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

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

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

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

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

Hence proved.  

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Updated on: 10-Oct-2022

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