The number of the + ive integral solutions of `tan^(-1)x+cos^(-1) "" y/(sqrt(1+y^2))=sin^(-1) (3/(sqrt10)) `
` " or " tan^(-1) x+ cot^(-1) y = tan^(-1) 3` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the number of positive integral solutions for the equation:

\[ \tan^{-1} x + \cos^{-1} \left(\frac{y}{\sqrt{1+y^2}}\right) = \sin^{-1} \left(\frac{3}{\sqrt{10}}\right) \]

or equivalently,

\[ \tan^{-1} x + \cot^{-1} y = \tan^{-1} 3 \]

### Step 1: Rewrite the equation

Using the identity \(\cot^{-1} y = \tan^{-1} \frac{1}{y}\), we can rewrite the equation as:

\[ \tan^{-1} x + \tan^{-1} \frac{1}{y} = \tan^{-1} 3 \]

### Step 2: Use the addition formula for \(\tan^{-1}\)

The addition formula for \(\tan^{-1}\) states that:

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

This is valid when \(ab < 1\). Here, we can apply this formula:

Let \(a = x\) and \(b = \frac{1}{y}\). Thus, we have:

\[ \tan^{-1} \left(\frac{x + \frac{1}{y}}{1 - x \cdot \frac{1}{y}}\right) = \tan^{-1} 3 \]

### Step 3: Set the arguments equal

Since the \(\tan^{-1}\) functions are equal, we can set the arguments equal to each other:

\[ \frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = 3 \]

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

Cross-multiplying gives us:

\[ x + \frac{1}{y} = 3 \left(1 - \frac{x}{y}\right) \]

Expanding this, we get:

\[ x + \frac{1}{y} = 3 - \frac{3x}{y} \]

### Step 5: Rearranging the equation

Rearranging gives us:

\[ x + \frac{1}{y} + \frac{3x}{y} = 3 \]

Combining the terms involving \(y\):

\[ x + \frac{1 + 3x}{y} = 3 \]

### Step 6: Isolate \(\frac{1 + 3x}{y}\)

Isolating \(\frac{1 + 3x}{y}\):

\[ \frac{1 + 3x}{y} = 3 - x \]

### Step 7: Solve for \(y\)

Now, we can express \(y\) in terms of \(x\):

\[ y = \frac{1 + 3x}{3 - x} \]

### Step 8: Determine the conditions for positive integral solutions

To find positive integral solutions, both \(x\) and \(y\) must be positive integers.

1. The denominator \(3 - x\) must be positive, which gives us \(x < 3\).
2. The possible positive integer values for \(x\) are \(1\) and \(2\).

### Step 9: Calculate \(y\) for each value of \(x\)

- For \(x = 1\):
\[ y = \frac{1 + 3(1)}{3 - 1} = \frac{4}{2} = 2 \]

- For \(x = 2\):
\[ y = \frac{1 + 3(2)}{3 - 2} = \frac{7}{1} = 7 \]

### Step 10: List the solutions

The pairs \((x, y)\) that satisfy the equation are:

1. \((1, 2)\)
2. \((2, 7)\)

### Conclusion

Thus, there are **2 positive integral solutions** to the equation.