Question

If $f\left(\theta \right)=(\sec \theta +\tan \theta -1)/(\tan \theta -\sec \theta +1)=\cos \theta /(1-\sin \theta )$, then the minimum value of $f\left(\theta \right)$ is

Answer 1

$f\left(\theta \right)=\frac{\cos \theta }{1-\sin \theta }$

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

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

$\Rightarrow \frac{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2})}{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}$

Dividing numerator & denominator,

$\Rightarrow \frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}$

$\Rightarrow \tan (\frac{\pi }{4}+\frac{\theta }{2})$

$\therefore$ Minimum value of $\tan$ is $0$.