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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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