If `sinx+cosx=1/5,0lexlepi,` then tan x is equal to

Options

A. `-4/3or, -3/4`
B. `3/4`
C. `4/5`
D. none of these

Explanation

To solve the problem where \( \sin x + \cos x = \frac{1}{5} \) and \( 0 \leq x \leq \pi \), we need to find the value of \( \tan x \). Here’s the step-by-step solution:

### Step 1: Square Both Sides
We start with the equation:
\[ \sin x + \cos x = \frac{1}{5} \]
Squaring both sides gives us:
\[ (\sin x + \cos x)^2 = \left(\frac{1}{5}\right)^2 \]
This expands to:
\[ \sin^2 x + \cos^2 x + 2\sin x \cos x = \frac{1}{25} \]

### Step 2: Use Pythagorean Identity
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can substitute:
\[ 1 + 2\sin x \cos x = \frac{1}{25} \]

### Step 3: Isolate \( \sin 2x \)
We know that \( 2\sin x \cos x = \sin 2x \). Therefore, we can rewrite the equation as:
\[ 1 + \sin 2x = \frac{1}{25} \]
Subtracting 1 from both sides gives:
\[ \sin 2x = \frac{1}{25} - 1 = \frac{1 - 25}{25} = -\frac{24}{25} \]

### Step 4: Solve for \( 2x \)
Now we have:
\[ \sin 2x = -\frac{24}{25} \]
Since \( \sin 2x \) is negative, \( 2x \) must be in the third or fourth quadrant. The general solution for \( \sin \theta = k \) is:
\[ \theta = \arcsin(k) + 2n\pi \quad \text{or} \quad \theta = \pi - \arcsin(k) + 2n\pi \]
For our case:
\[ 2x = \arcsin\left(-\frac{24}{25}\right) + 2n\pi \quad \text{or} \quad 2x = \pi - \arcsin\left(-\frac{24}{25}\right) + 2n\pi \]
However, since \( 0 \leq x \leq \pi \), we will consider only the principal values.

### Step 5: Calculate \( x \)
Calculating \( 2x \):
\[ 2x = \arcsin\left(-\frac{24}{25}\right) + 2\pi \quad \text{(not valid since } 2x \text{ must be } \leq 2\pi\text{)} \]
or
\[ 2x = \pi + \arcsin\left(-\frac{24}{25}\right) \]
This gives us:
\[ x = \frac{\pi}{2} + \frac{1}{2}\arcsin\left(-\frac{24}{25}\right) \]

### Step 6: Find \( \tan x \)
Using the identity \( \tan x = \frac{\sin x}{\cos x} \), we can find \( \tan x \) using the values of \( \sin x \) and \( \cos x \). However, we can also express \( \tan x \) in terms of \( \sin 2x \):
\[ \tan x = \frac{\sin 2x}{1 + \cos 2x} \]
We need to find \( \cos 2x \):
\[ \cos 2x = \sqrt{1 - \sin^2 2x} = \sqrt{1 - \left(-\frac{24}{25}\right)^2} = \sqrt{1 - \frac{576}{625}} = \sqrt{\frac{49}{625}} = \frac{7}{25} \]
Thus:
\[ \tan x = \frac{-\frac{24}{25}}{1 + \frac{7}{25}} = \frac{-\frac{24}{25}}{\frac{32}{25}} = -\frac{24}{32} = -\frac{3}{4} \]

### Final Answer
Thus, the value of \( \tan x \) is:
\[ \tan x = -\frac{3}{4} \]