Solve `tan^-1(1/(1+2x))+tan^-1(1/(1+4x))=tan^-1(2/x^2)`

Answer

To solve the equation \( \tan^{-1}\left(\frac{1}{1+2x}\right) + \tan^{-1}\left(\frac{1}{1+4x}\right) = \tan^{-1}\left(\frac{2}{x^2}\right) \), we will use the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

### Step 1: Apply the formula

Let \( a = \frac{1}{1+2x} \) and \( b = \frac{1}{1+4x} \). Then we can write:

\[ \tan^{-1}\left(\frac{1}{1+2x}\right) + \tan^{-1}\left(\frac{1}{1+4x}\right) = \tan^{-1}\left(\frac{\frac{1}{1+2x} + \frac{1}{1+4x}}{1 - \frac{1}{1+2x} \cdot \frac{1}{1+4x}}\right) \]

### Step 2: Simplify the numerator

To simplify the numerator:

\[ \frac{1}{1+2x} + \frac{1}{1+4x} = \frac{(1+4x) + (1+2x)}{(1+2x)(1+4x)} = \frac{2 + 6x}{(1+2x)(1+4x)} \]

### Step 3: Simplify the denominator

Now simplify the denominator:

\[ 1 - \frac{1}{(1+2x)(1+4x)} = \frac{(1+2x)(1+4x) - 1}{(1+2x)(1+4x)} \]

Calculating \( (1+2x)(1+4x) - 1 \):

\[ (1 + 2x)(1 + 4x) = 1 + 4x + 2x + 8x^2 = 1 + 6x + 8x^2 \]
\[ (1 + 6x + 8x^2) - 1 = 6x + 8x^2 \]

So, the denominator becomes:

\[ \frac{6x + 8x^2}{(1+2x)(1+4x)} \]

### Step 4: Combine the results

Now we can substitute back into our equation:

\[ \tan^{-1}\left(\frac{\frac{2 + 6x}{(1+2x)(1+4x)}}{\frac{6x + 8x^2}{(1+2x)(1+4x)}}\right) = \tan^{-1}\left(\frac{2 + 6x}{6x + 8x^2}\right) \]

### Step 5: Set the two sides equal

Now we have:

\[ \tan^{-1}\left(\frac{2 + 6x}{6x + 8x^2}\right) = \tan^{-1}\left(\frac{2}{x^2}\right) \]

### Step 6: Remove the inverse tangent

Since the tangent function is one-to-one, we can equate the arguments:

\[ \frac{2 + 6x}{6x + 8x^2} = \frac{2}{x^2} \]

### Step 7: Cross-multiply

Cross-multiplying gives:

\[ (2 + 6x)x^2 = 2(6x + 8x^2) \]

### Step 8: Expand both sides

Expanding both sides:

\[ 2x^2 + 6x^3 = 12x + 16x^2 \]

### Step 9: Rearrange the equation

Rearranging gives:

\[ 6x^3 - 14x^2 - 12x = 0 \]

### Step 10: Factor out common terms

Factoring out \( 2x \):

\[ 2x(3x^2 - 7x - 6) = 0 \]

### Step 11: Solve the quadratic

Setting \( 2x = 0 \) gives \( x = 0 \). Now solve the quadratic \( 3x^2 - 7x - 6 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3} \]

Calculating the discriminant:

\[ 49 + 72 = 121 \]

So:

\[ x = \frac{7 \pm 11}{6} \]

This gives:

\[ x = 3 \quad \text{and} \quad x = -\frac{2}{3} \]

### Final Solutions

Thus, the solutions are:

\[ x = 0, \quad x = 3, \quad x = -\frac{2}{3} \]