Question

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

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

Answer 1

The correct option is B

$-\frac{\left(4+\sqrt{7}\right)}{3}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\cos \left(x\right)+\sin \left(x\right)=\frac{1}{2}$

Squaring both side.

$\Rightarrow$${\left(\cos \left(x\right)+\sin \left(x\right)\right)}^2={\left(\frac{1}{2}\right)}^2$

$\Rightarrow$${\cos }^2\left(x\right)+{\sin }^2\left(x\right)+2\sin \left(x\right)\cos \left(x\right)=\frac{1}{4}$

$\Rightarrow$ $\sin \left(2x\right)=\frac{1}{4}-1$

$\Rightarrow$ $\frac{2\tan \left(x\right)}{1+{\tan }^2\left(x\right)}=\left(-\frac{3}{4}\right)$

$\Rightarrow$ $3{\tan }^2\left(x\right)+8\tan \left(x\right)+3=0$

Step-2: Solve the quadratic equation by using formula $\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}$

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

$\Rightarrow$ $\tan \left(x\right)=\frac{-8\pm 2\sqrt{7}}{2\times 3}$

$\Rightarrow$$\tan \left(x\right)=\left(-\left(\frac{4+\sqrt{7}}{3}\right),\left(\frac{-4+\sqrt{7}}{3}\right)\right)$

Hence, correct answer is option $\left(B\right)$.