Question

If $\cos x+\sin x=\frac{1}{2}$, where $x\in (0,\pi )$, then the maximum possible value of $\tan x$ is

• A. $\frac{-4-\sqrt{7}}{3}$
• B. $\frac{4+\sqrt{7}}{3}$
• C. $\frac{-4+\sqrt{7}}{3}$
• D. $\frac{4-\sqrt{7}}{3}$

Answer 1

The correct option is C $\frac{-4+\sqrt{7}}{3}$

$\cos x+\sin x=\frac{1}{2}$
Squaring both sides,
$\Rightarrow 1+\sin 2x=\frac{1}{4}$
$\Rightarrow \sin 2x=-\frac{3}{4}$

When $x\in (0,\frac{\pi }{2})\to 2x\in (0,\pi )$
Then $\sin 2x\to \text{positive}$
When $x\in (\frac{\pi }{2},\pi )\to 2x\in (\pi ,2\pi )$
Then $\sin 2x\to \text{negative}$
So,
$x\in (\frac{\pi }{2},\pi )$
Now,
$\sin 2x=-\frac{3}{4}\Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}=-\frac{3}{4}$
$\Rightarrow 3{\tan }^{2}x+8\tan x+3=0$
$\Rightarrow \tan x=\frac{-8\pm \sqrt{64-36}}{6}\Rightarrow \tan x=\frac{-4\pm \sqrt{7}}{3}$
As both values are negative, so both are acceptable.
Therefore the maximum possible value is,
$\tan x=\frac{-4+\sqrt{7}}{3}$