Question

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

• A. $\tan \left(\frac{\theta }{2}-\frac{\pi }{4}\right)$
• B. $\tan \left(-\frac{\pi }{4}-\frac{\theta }{2}\right)$
• C. $\tan \left(\frac{\pi }{4}-\frac{\theta }{2}\right)$
• D. $\tan \left(\frac{\pi }{4}+\frac{\theta }{2}\right)$

Answer 1

Answer: (C)

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