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Q. The value of $\frac{\cos\theta}{1+ \sin\theta}$ is equal to

VITEEEVITEEE 2010Trigonometric Functions

Solution:

$\frac{\cos\theta}{1+ \sin\theta} = \frac{ \sin\left(\frac{\pi}{2} - \theta\right)}{1 + \cos \left(\frac{\pi}{2} - \theta\right)} $
$= \frac{2 \sin\left( \frac{\pi}{4} - \frac{\theta}{2}\right)\cos \left(\frac{\pi}{4} - \frac{\theta}{2}\right)}{2 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)} $
$ = \frac{\sin\left(\frac{\pi}{4} - \frac{\theta}{2}\right)}{\cos\left(\frac{\pi}{4} - \frac{\theta}{2}\right)}$
$= \tan \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$
Alternate $\frac{\cos \theta}{1+\sin \theta}$
$=\frac{\left(\cos ^{2} \theta / 2-\sin ^{2} \theta / 2\right)}{\left(\sin ^{2} \theta / 2+\cos ^{2} \theta / 2\right)+2 \sin \theta / 2+\cos \theta / 2}$
$=\frac{(\cos \theta / 2+\sin \theta / 2)(\cos \theta / 2-\sin \theta / 2)}{(\cos \theta / 2+\sin \theta / 2)^{2}}$
$=\frac{1-\tan \theta / 2}{1+\tan \theta / 2}$
$=\frac{\tan \pi / 4-\tan \theta / 2}{1+\tan \pi / 4 \cdot \tan \theta / 2}$
$=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$