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Prove the following identities:
\( \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta} \)
To do:
We have to prove that \( \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta} \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$
$=\frac{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$
$=\frac{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$
$=\frac{(\sec \theta-\tan \theta)[\sec \theta+\tan \theta+1]}{1+\sec \theta+\tan \theta}$
$=\sec \theta-\tan \theta$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
$=\frac{1-\sin \theta}{\cos \theta}$
Hence proved.