Question

If $\sin x+\cos x=\frac{1}{5}$ then $\tan \left(2x\right)$ is

• A. $\frac{25}{17}$
• B. $\frac{7}{25}$
• C. $\frac{25}{7}$
• D. $\frac{24}{7}$

Answer 1

The correct option is D

$\frac{24}{7}$

Explanation for the correct option

Step 1: Given the equation

$\therefore \sin x+\cos x=\frac{1}{5}$

Squaring both sides

$\begin{array}{rcl}& \Rightarrow & {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=\frac{1}{25}\end{array}$

From the identity formula ${\sin }^{2}x+{\cos }^{2}x=1$

$\begin{array}{rcl}\therefore 1+\sin 2x& =& \frac{1}{25}\\ & \Rightarrow & \sin 2x=\frac{1}{25}-1\\ & \Rightarrow & \sin 2x=\frac{1-25}{25}\\ & \Rightarrow & \sin 2x=-\frac{24}{25}\end{array}$

Step 3: Using the identity formula ${\sin }^{2}x+{\cos }^{2}x=1$

Therefore also ${\sin }^{2}2x+{\cos }^{2}2x=1$for finding cos2x

$\begin{array}{rcl}\therefore {\cos }^{2}2x& =& 1-{\sin }^{2}2x\\ & \Rightarrow & \cos 2x=\sqrt{1-{\sin }^{2}2x}\\ & \Rightarrow & \cos 2x=\sqrt{1-{\left(\frac{\left(-24\right)}{25}\right)}^2}\\ & \Rightarrow & \cos 2x=\sqrt{1-\frac{576}{625}}\\ & \Rightarrow & \cos 2x=\sqrt{\frac{49}{625}}\\ & \Rightarrow & \cos 2x=+\frac{7}{25}or-\frac{7}{25}\end{array}$

$\because$$\sin 2x\mathrm{and}\cos 2x$ are given therefore we can find the value of $\tan 2x$

When $\cos 2x=\left(\frac{7}{25}\right)$

$$\begin{gathered} \therefore \tan 2x=\frac{\sin 2x}{\cos 2x} \\ \Rightarrow \tan 2x=\frac{\left({\displaystyle \frac{-24}{25}}\right)}{{\displaystyle \left(\frac{7}{25}\right)}} \\ \Rightarrow \tan 2x=\left(\frac{-24}{7}\right) \end{gathered}$$

When $\cos 2x=\left(\frac{-7}{25}\right)$

$$\begin{gathered} \therefore \tan 2x=\frac{\sin 2x}{\cos 2x} \\ \Rightarrow \tan 2x=\frac{\left({\displaystyle \frac{-24}{25}}\right)}{{\displaystyle \left(\frac{-7}{25}\right)}} \\ \Rightarrow \tan 2x=\left(\frac{24}{7}\right) \end{gathered}$$

Hence option (D) is the correct answer.