Question

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

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

Answer 1

Answer: (A)

$\frac{4-\sqrt{7}}{3}$

$\cos x+\sin x=\frac{1}{2}$ $x\in (\pi ,2\pi )$

divide by $\cos x$

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

$\Rightarrow 1+\tan x=\frac{1}{2}\sec x$ [we know ${\sec }^{2}x-{\tan }^{2}x=1$]

$\Rightarrow 1+\tan x=\frac{1}{2}\sqrt{1+{\tan }^{2}x}$

Let $\tan x=t$

$1+t=\frac{\sqrt{1+t^2}}{2}$

$\left(2\right(1+t){)}^2=1+t^2$

$\Rightarrow 3t^2+8t+3=0$

$\therefore t=\frac{-8\pm \sqrt{8^2-4\left(3\right)\left(3\right)}}{2\times 3}$

$\Rightarrow \frac{-8\pm 2\sqrt{7}}{6}$

$\Rightarrow t=\frac{-4\pm \sqrt{7}}{3}$

$\therefore \tan x=\frac{-4\pm \sqrt{7}}{3}$ but $x\in (\pi ,2\pi )$

$\therefore \tan x=\frac{-4+\sqrt{7}}{3}$.