Question

$\tan x+\sec x=2\cos x,0<x<\frac{\pi }{2},thenx=..............$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{\pi }{6}$

we have, $\tan x+\sec x=2\cos x$

$\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$

$\Rightarrow \sin x+1=2{\cos }^{2}x$

$\Rightarrow \sin x+1=2-2{\sin }^{2}x$

$\Rightarrow 2{\sin }^{2}x+\sin x-1=0$

$\sin x=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm 3}{4}$

$\sin x=-1$ $\sin x=\frac{1}{2}$

$x=\frac{3\pi }{2}$ but $0<x<\frac{\pi }{2}$

So, $\sin x=1/2$ $\Rightarrow$ $x=\frac{5\pi }{6}$