Sometimes we are just concerned with finding integral solutions to equations. Consider the equation
`tan^(-1).(1)/m+tan^(-1).(1)/n=tan^(-1).(1)/lambda`, where `m,n, lambda in N`
If `lambda` is such that `lambda^2=1` is a prime, then how many solutions (m,n) are there for the equation?

Options

A. One
B. Two
C. Four
D. Infinite

Answer

To solve the equation \( \tan^{-1}\left(\frac{1}{m}\right) + \tan^{-1}\left(\frac{1}{n}\right) = \tan^{-1}\left(\frac{1}{\lambda}\right) \), where \( m, n, \lambda \in \mathbb{N} \) and \( \lambda^2 = 1 \) is a prime, we can follow these steps:

### Step 1: Understand the values of \( \lambda \)
Since \( \lambda^2 = 1 \) and \( \lambda \) is a natural number, the possible values for \( \lambda \) are \( 1 \) and \( -1 \). However, since \( \lambda \) must be a natural number, we only consider \( \lambda = 1 \).

### Step 2: Use the addition formula for inverse tangent
We use the formula for the sum of two inverse tangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]
Setting \( x = \frac{1}{m} \) and \( y = \frac{1}{n} \), we have:
\[ \tan^{-1}\left(\frac{1}{m}\right) + \tan^{-1}\left(\frac{1}{n}\right) = \tan^{-1}\left(\frac{\frac{1}{m} + \frac{1}{n}}{1 - \frac{1}{mn}}\right) \]

### Step 3: Simplify the expression
The expression simplifies to:
\[ \tan^{-1}\left(\frac{\frac{n + m}{mn}}{1 - \frac{1}{mn}}\right) = \tan^{-1}\left(\frac{m + n}{mn - 1}\right) \]

### Step 4: Set the equation equal to \( \tan^{-1}(1) \)
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ \tan^{-1}\left(\frac{m+n}{mn-1}\right) = \tan^{-1}(1) \]
This implies:
\[ \frac{m+n}{mn-1} = 1 \]

### Step 5: Solve the equation
From the equation \( \frac{m+n}{mn-1} = 1 \), we can cross-multiply:
\[ m + n = mn - 1 \]
Rearranging gives us:
\[ mn - m - n - 1 = 0 \]

### Step 6: Rearranging into a standard form
We can rearrange this into:
\[ mn - m - n = 1 \]
This can be factored as:
\[ (m-1)(n-1) = 2 \]

### Step 7: Finding integral solutions
Now we need to find the pairs \((m-1, n-1)\) that multiply to \(2\):
- The factor pairs of \(2\) are \((1, 2)\) and \((2, 1)\).
- This gives us the pairs:
1. \( (m-1, n-1) = (1, 2) \) leads to \( m = 2, n = 3 \)
2. \( (m-1, n-1) = (2, 1) \) leads to \( m = 3, n = 2 \)

Thus, the integral solutions \((m, n)\) are \((2, 3)\) and \((3, 2)\).

### Conclusion
Therefore, there are **2 integral solutions** for the equation.

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