If `tan x=(a)/(b)` and `tan 2x=(b)/(a+b)` find the smallest positive value of x.

Explanation

To solve the problem, we start with the given equations:

1. \( \tan x = \frac{a}{b} \)
2. \( \tan 2x = \frac{b}{a+b} \)

We will use the double angle formula for tangent, which is:

\[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \]

### Step 1: Substitute \( \tan x \) into the double angle formula

Substituting \( \tan x = \frac{a}{b} \) into the double angle formula:

\[ \tan 2x = \frac{2 \cdot \frac{a}{b}}{1 - \left(\frac{a}{b}\right)^2} \]

### Step 2: Simplify the expression

Now, simplify the expression:

\[ \tan 2x = \frac{\frac{2a}{b}}{1 - \frac{a^2}{b^2}} = \frac{\frac{2a}{b}}{\frac{b^2 - a^2}{b^2}} = \frac{2a \cdot b^2}{b(b^2 - a^2)} = \frac{2ab}{b^2 - a^2} \]

### Step 3: Set the two expressions for \( \tan 2x \) equal

Now we set the two expressions for \( \tan 2x \) equal to each other:

\[ \frac{2ab}{b^2 - a^2} = \frac{b}{a + b} \]

### Step 4: Cross-multiply to eliminate the fractions

Cross-multiplying gives:

\[ 2ab(a + b) = b(b^2 - a^2) \]

### Step 5: Expand both sides

Expanding both sides results in:

\[ 2a^2b + 2ab^2 = b^3 - ab^2 \]

### Step 6: Rearrange the equation

Rearranging the equation gives:

\[ 2a^2b + 2ab^2 + ab^2 - b^3 = 0 \]

This simplifies to:

\[ 2a^2b + 3ab^2 - b^3 = 0 \]

### Step 7: Factor out \( b \)

Factoring out \( b \) from the equation:

\[ b(2a^2 + 3ab - b^2) = 0 \]

Since \( b \neq 0 \), we can set the quadratic equation to zero:

\[ 2a^2 + 3ab - b^2 = 0 \]

### Step 8: Solve the quadratic equation for \( a \)

Using the quadratic formula \( a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):

Here, \( A = 2, B = 3b, C = -b^2 \):

\[ a = \frac{-3b \pm \sqrt{(3b)^2 - 4 \cdot 2 \cdot (-b^2)}}{2 \cdot 2} \]
\[ = \frac{-3b \pm \sqrt{9b^2 + 8b^2}}{4} \]
\[ = \frac{-3b \pm \sqrt{17b^2}}{4} \]
\[ = \frac{-3b \pm b\sqrt{17}}{4} \]

### Step 9: Find \( \tan x \)

Since \( \tan x = \frac{a}{b} \), we have:

\[ \tan x = \frac{-3 \pm \sqrt{17}}{4} \]

### Step 10: Determine the smallest positive value of \( x \)

We take the positive value:

\[ \tan x = \frac{-3 + \sqrt{17}}{4} \]

Now, to find \( x \):

\[ x = \tan^{-1}\left(\frac{-3 + \sqrt{17}}{4}\right) \]

### Final Answer

Thus, the smallest positive value of \( x \) is:

\[ x = \tan^{-1}\left(\frac{-3 + \sqrt{17}}{4}\right) \]

---