Question

If the equation $\sec \theta +cosec\theta =c$ has two real roots between $0$ and $2\pi$ then the least integer which $c^2$ cannot exceed must be:

Answer 1

Given, $c=\frac{1}{\cos \theta }+\frac{1}{\sin \theta }$

$=\frac{\sin \theta +\cos \theta }{\cos \theta .\sin \theta }$

$=\frac{\sqrt{2}\sin (\theta +{45}^0)}{\frac{\sin 2\theta }{2}}$

$=\frac{2\sqrt{2}\sin (\theta +{45}^0)}{\sin 2\theta }$

Hence

$c=\frac{2\sqrt{2}\sin (\theta +{45}^0)}{\sin 2\theta }$

Now consider

$y=\frac{\sin (\theta +{45}^0)}{\sin 2\theta }$

$y^{\prime }=\frac{\cos (\theta +{45}^0)\sin 2\theta -2\sin (\theta +{45}^0).\cos 2\theta }{{\sin }^{2}2\theta }$$=0$

$\tan ({45}^0+\theta )=\frac{1}{2}\tan \left(2\theta \right)$

$\frac{1+\tan \theta }{1-\tan \theta }=\frac{\tan \theta }{1-{\tan }^2\theta }$

$(1+\tan \theta )(1-{\tan }^2\theta )=\tan \theta (1-\tan \theta )$

$(1-\tan \theta )\left(\right(1+\tan \theta {)}^2-\tan \theta )=0$

$(1-\tan \theta )(1+\tan \theta +{\tan }^2\theta )=0$

Now, $1+\tan \theta +{\tan }^2\theta \ne 0$

Hence

$1-\tan \theta =0$

$\tan \theta =1$

$\theta ={45}^0,{225}^0$

Hence $f\left(\theta \right)$ attains a maximum value at $\theta ={45}^0$

Hence

$c\le 2\sqrt{2}$

$c^2\le 8$.