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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.