Question

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

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

Answer 1

The correct option is B

$-\frac{\left(4–\sqrt{7}\right)}{3}$

Step 1. Find the value of $\mathit{t}\mathit{a}\mathit{n}\mathit{x}$:

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

Squaring on both sides,

${\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x=\frac{1}{4}$

$\Rightarrow$ $1+\sin 2x=\frac{1}{4}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{;}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\right]$

$\Rightarrow$ $\sin 2x=\frac{1}{4}–1$

$\Rightarrow$ $\frac{2\tan x}{(1+{\tan }^{2}x)}=-\frac{3}{4}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }}{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}\right]$

$\Rightarrow$ $8\tan x=-3–3{\tan }^{2}x$

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

Step 2. By using quadratic formula, we get

$\begin{array}{rcl}\tan x& =& \frac{-8\pm \sqrt{(64–36)}}{2\left(3\right)}\\ & =& \frac{-8\pm \sqrt{28}}{6}\\ & =& \frac{-8\pm 2\sqrt{7}}{6}\\ & =& \frac{-4\pm \sqrt{7}}{3}\end{array}$

$\begin{array}{l}\frac{-4–\sqrt{7}}{3}\end{array}$ is not possible as $0<x<\pi$

$\mathbf{\therefore }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\frac{\mathbf{(}\mathbf{4}\mathbf{}\mathbf{-}\mathbf{}\sqrt{\mathbf{7}}\mathbf{)}}{\mathbf{3}}$

Hence, Option ‘B’ is Correct.