Question

If $\alpha ,\beta$ are the roots of the equation $\tan x+\sec x=2\cos x$ where $x\in [0,2\pi )$, then the value of $|\alpha -\beta |$ is

• A. $\frac{2\pi }{3}$
• B. $\frac{5\pi }{6}$
• C. $\frac{2\pi }{5}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{2\pi }{3}$

$\tan x+\sec x=2\cos x$
$\Rightarrow \sin x+1=2{\cos }^{2}x,x\ne (2n+1)\frac{\pi }{2}$
$\Rightarrow 2{\sin }^{2}x+\sin x-1=0$
$\Rightarrow (2\sin x-1)(\sin x+1)=0$
$\Rightarrow \sin x=\frac{1}{2},\sin x=-1$

$\sin x=-1$ not possible as $x\ne (2n+1)\frac{\pi }{2}$
$\therefore \sin x=\frac{1}{2}$
$\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}$
$\therefore |\alpha -\beta |=\frac{2\pi }{3}$