Prove that `tan^(-1)1+tan^(-1)2+tan^(-1)3 =pi`

Answer

To prove that \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \), we can follow these steps:

### Step 1: Set Variables
Let:
\[ x = \tan^{-1}(1), \quad y = \tan^{-1}(2), \quad z = \tan^{-1}(3) \]

### Step 2: Express Tangents
From the definitions of the inverse tangent:
\[ \tan(x) = 1, \quad \tan(y) = 2, \quad \tan(z) = 3 \]

### Step 3: Use the Tangent Addition Formula
We will use the tangent addition formula:
\[ \tan(x + y + z) = \frac{\tan(x) + \tan(y) + \tan(z) - \tan(x)\tan(y)\tan(z)}{1 - \tan(x)\tan(y) - \tan(y)\tan(z) - \tan(z)\tan(x)} \]

### Step 4: Substitute Values
Substituting the values of \( \tan(x) \), \( \tan(y) \), and \( \tan(z) \):
\[ \tan(x + y + z) = \frac{1 + 2 + 3 - (1)(2)(3)}{1 - (1)(2) - (2)(3) - (3)(1)} \]

### Step 5: Simplify the Numerator
Calculating the numerator:
\[ 1 + 2 + 3 = 6 \]
\[ (1)(2)(3) = 6 \]
Thus, the numerator becomes:
\[ 6 - 6 = 0 \]

### Step 6: Simplify the Denominator
Calculating the denominator:
\[ 1 - (1)(2) - (2)(3) - (3)(1) = 1 - 2 - 6 - 3 = 1 - 11 = -10 \]

### Step 7: Final Calculation
Now substituting back into the tangent formula:
\[ \tan(x + y + z) = \frac{0}{-10} = 0 \]

### Step 8: Conclusion
Since \( \tan(x + y + z) = 0 \), it implies:
\[ x + y + z = n\pi \quad \text{for some integer } n \]
Given that \( x = \tan^{-1}(1) \), \( y = \tan^{-1}(2) \), and \( z = \tan^{-1}(3) \) are all positive angles, we conclude:
\[ x + y + z = \pi \]

Thus, we have:
\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \]

### Final Statement
Therefore, we have proved that:
\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \]

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