If `sinx+cosx=1/5`, then find the value of `tan 2x`.

Answer

To solve the problem where \( \sin x + \cos x = \frac{1}{5} \) and we need to find the value of \( \tan 2x \), we can follow these steps:

### Step 1: Express \( \cos x \) in terms of \( \sin x \)
From the equation \( \sin x + \cos x = \frac{1}{5} \), we can express \( \cos x \) as:
\[ \cos x = \frac{1}{5} - \sin x \]

### Step 2: Use the Pythagorean identity
We know that \( \sin^2 x + \cos^2 x = 1 \). Substituting \( \cos x \) from Step 1:
\[ \sin^2 x + \left(\frac{1}{5} - \sin x\right)^2 = 1 \]
Expanding the square:
\[ \sin^2 x + \left(\frac{1}{25} - \frac{2}{5}\sin x + \sin^2 x\right) = 1 \]
Combining like terms:
\[ 2\sin^2 x - \frac{2}{5}\sin x + \frac{1}{25} = 1 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ 2\sin^2 x - \frac{2}{5}\sin x + \frac{1}{25} - 1 = 0 \]
Simplifying further:
\[ 2\sin^2 x - \frac{2}{5}\sin x - \frac{24}{25} = 0 \]

### Step 4: Multiply through by 25 to eliminate fractions
\[ 50\sin^2 x - 10\sin x - 24 = 0 \]

### Step 5: Factor or use the quadratic formula
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 50, b = -10, c = -24 \):
\[ \sin x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 50 \cdot (-24)}}{2 \cdot 50} \]
Calculating the discriminant:
\[ \sqrt{100 + 4800} = \sqrt{4900} = 70 \]
Thus:
\[ \sin x = \frac{10 \pm 70}{100} \]
Calculating the two possible values:
1. \( \sin x = \frac{80}{100} = \frac{4}{5} \)
2. \( \sin x = \frac{-60}{100} = -\frac{3}{5} \)

### Step 6: Find corresponding \( \cos x \) values
Using \( \cos x = \frac{1}{5} - \sin x \):
1. For \( \sin x = \frac{4}{5} \):
\[ \cos x = \frac{1}{5} - \frac{4}{5} = -\frac{3}{5} \]
2. For \( \sin x = -\frac{3}{5} \):
\[ \cos x = \frac{1}{5} + \frac{3}{5} = \frac{4}{5} \]

### Step 7: Calculate \( \tan x \)
Using \( \tan x = \frac{\sin x}{\cos x} \):
1. For \( \sin x = \frac{4}{5}, \cos x = -\frac{3}{5} \):
\[ \tan x = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \]
2. For \( \sin x = -\frac{3}{5}, \cos x = \frac{4}{5} \):
\[ \tan x = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \]

### Step 8: Calculate \( \tan 2x \)
Using the formula \( \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \):
1. For \( \tan x = -\frac{4}{3} \):
\[ \tan 2x = \frac{2 \cdot -\frac{4}{3}}{1 - \left(-\frac{4}{3}\right)^2} = \frac{-\frac{8}{3}}{1 - \frac{16}{9}} = \frac{-\frac{8}{3}}{-\frac{7}{9}} = \frac{8 \cdot 9}{3 \cdot 7} = \frac{24}{7} \]
2. For \( \tan x = -\frac{3}{4} \):
\[ \tan 2x = \frac{2 \cdot -\frac{3}{4}}{1 - \left(-\frac{3}{4}\right)^2} = \frac{-\frac{3}{2}}{1 - \frac{9}{16}} = \frac{-\frac{3}{2}}{\frac{7}{16}} = \frac{-3 \cdot 16}{2 \cdot 7} = \frac{-24}{7} \]

### Final Result
Thus, the values of \( \tan 2x \) are \( \frac{24}{7} \) and \( -\frac{24}{7} \).