Question

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

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

Answer 1

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


$cosx+sinx=\frac{1}{2}\Rightarrow (cosx+sinx{)}^2=\frac{1}{4}\Rightarrow 1+sin2x=\frac{1}{4}\Rightarrow sin2x=-\frac{3}{4}$
$\therefore \text{x is obtuse and}\frac{2tanx}{1+ta{n}^{2}x}=-\frac{3}{4}\Rightarrow 3ta{n}^{2}x+8tanx+3=0$
$\Rightarrow tanx=\frac{-8\pm \sqrt{64-36}}{6}=\frac{-4\pm \sqrt{7}}{3}$