Find the number of positive integral solution of the equation `tan^(-1)x+cos^(-1)(y/(sqrt(1-y^2)))=sin^(-1)(3/(sqrt(10)))`

Options

A. `(1,2 )` & `(2,7)`
B. `(1,-2 )` & `(2,7)`
C. `(1,2 )` & `(-2,7)`
D. `(1,2 )` & `(2,-7)`

Explanation

To find the number of positive integral solutions of the equation

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

we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities.

We know that

\[ \cos^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) = \tan^{-1}\left(\frac{1}{y}\right), \]

because if we let \( \theta = \cos^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \), then \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sqrt{1-y^2}}{y} \), which gives us \( \tan^{-1}\left(\frac{1}{y}\right) \).

Thus, we can rewrite the equation as:

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

### Step 2: Convert \(\sin^{-1}\left(\frac{3}{\sqrt{10}}\right)\) to \(\tan^{-1}\).

Using the identity for \(\sin^{-1}(a)\):

\[ \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right), \]

we find:

\[ \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan^{-1}\left(\frac{3}{\sqrt{10 - 9}}\right) = \tan^{-1}(3). \]

### Step 3: Substitute back into the equation.

Now, we have:

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

### Step 4: Use the tangent addition formula.

Using the formula for the tangent of a sum, we have:

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

Setting this equal to \(\tan^{-1}(3)\), we get:

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

### Step 5: Cross-multiply and simplify.

Cross-multiplying gives:

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

This simplifies to:

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

Rearranging gives:

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

### Step 6: Combine terms.

This can be rewritten as:

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

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

Rearranging gives:

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

Multiplying through by \(y\) to eliminate the fraction gives:

\[ xy + 3x = 3y - 1. \]

Rearranging leads to:

\[ xy - 3y + 3x + 1 = 0. \]

### Step 8: Factor the equation.

This is a linear equation in \(y\):

\[ y(x - 3) = -3x - 1. \]

Thus,

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

### Step 9: Find positive integer solutions.

For \(y\) to be positive, the numerator and denominator must have the same sign. We analyze the cases:

1. **Case 1:** \(x - 3 > 0 \Rightarrow x > 3\) and \(-3x - 1 < 0 \Rightarrow x < -\frac{1}{3}\) (not possible since \(x\) is positive).
2. **Case 2:** \(x - 3 < 0 \Rightarrow x < 3\) and \(-3x - 1 > 0 \Rightarrow x < -\frac{1}{3}\) (not applicable).

### Step 10: Check integer values.

We can check integer values for \(x = 1\) and \(x = 2\):

- For \(x = 1\):

\[ y = \frac{-3(1) - 1}{1 - 3} = \frac{-4}{-2} = 2. \]

- For \(x = 2\):

\[ y = \frac{-3(2) - 1}{2 - 3} = \frac{-7}{-1} = 7. \]

Thus, the pairs \((x, y)\) are \((1, 2)\) and \((2, 7)\).

### Conclusion

The number of positive integral solutions is **2**: \((1, 2)\) and \((2, 7)\).

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