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

To do:

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

Solution:

We know that,

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

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

$\tan \theta=\frac{\sin \theta}{\cos \theta}$........(iii)

Therefore,

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

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

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

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

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

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

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

$=(\sec \theta-\tan \theta)^2$          (From (ii) and (iii))

Hence proved.

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Updated on: 10-Oct-2022

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