Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B $-\frac{(4-\sqrt{7})}{3}$
C $-\frac{(4+\sqrt{7})}{3}$

$\cos x+\sin x=\frac{1}{2}$ then $\tan x$ is?

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

Dividing by $\cos x$

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

$=1+\tan x=\frac{\sec x}{2}$

Squaring on both sides,

$(1+\tan x{)}^2=\frac{se{c}^{2}x}{4}$

$\Rightarrow 4(1+{\tan }^{2}x+2\tan x)={\sec }^{2}x$

$\Rightarrow 4+4{\tan }^{2}x+8\tan x=1+{\tan }^{2}x$ .......... $[{\sec }^2\theta =1+{\tan }^2\theta ]$

$\Rightarrow 3{\tan }^{2}x+8\tan x+3=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow \tan x=\frac{-8\pm \sqrt{8^2-4\times 3\times 3}}{2\times 3}$

$=\frac{-8\pm \sqrt{64-36}}{6}$

$=\frac{-8\pm \sqrt{28}}{6}$

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

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