Question

If $0<x<\pi$ and $\cos x+\sin x=\frac{1}{2}$ , then $\tan x$ is

• A. $\frac{(1-\sqrt{7})}{4}$
• B. $\frac{(4-\sqrt{7})}{4}$
• C. $-\frac{(4+\sqrt{7})}{3}$
• D. $\frac{(1+\sqrt{7})}{4}$

Answer 1

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

Given $\cos x+\sin x=\frac{1}{2}$
Squaring both sides ,we get
$\Rightarrow 1+\sin 2x=\frac{1}{4}$
$\Rightarrow \sin 2x=-\frac{3}{4}$
$\therefore x$ is obtuse
and $\frac{2\tan x}{1+{\tan }^{2}x}=-\frac{3}{4}$
$\Rightarrow 3{\tan }^{2}x+8\tan x+3=0$
$\Rightarrow \tan x=\frac{-8\pm \sqrt{64-36}}{6}$
$=\frac{-4\pm \sqrt{7}}{3}$
$\because \tan x<0\Rightarrow \tan x=-\frac{4-\sqrt{7}}{8}$