The set of lines `x tan^(- 1)a+ysin^(- 1)(1/(sqrt(1+a^2)))+2=0` where ` ain (0,1) ` are concurrent at (a) `(1/pi,1/pi)` (b) `(-4/pi,-4/pi)` (c) `(pi,pi)` (d) none of these

Explanation

To solve the problem, we need to determine the point at which the set of lines given by the equation

\[ x \tan^{-1} a + y \sin^{-1} \left(\frac{1}{\sqrt{1 + a^2}}\right) + 2 = 0 \]

is concurrent for \( a \in (0, 1) \).

### Step 1: Rewrite the equation
We start with the equation of the lines:

\[ x \tan^{-1} a + y \sin^{-1} \left(\frac{1}{\sqrt{1 + a^2}}\right) + 2 = 0 \]

### Step 2: Substitute variables
Let \( \sin^{-1} \left(\frac{1}{\sqrt{1 + a^2}}\right) = \theta \). This implies:

\[ \frac{1}{\sqrt{1 + a^2}} = \sin \theta \]

### Step 3: Analyze the triangle
From the definition of sine, we can form a right triangle where:

- The opposite side (perpendicular) is 1.
- The hypotenuse is \( \sqrt{1 + a^2} \).

### Step 4: Relate \( a \) and \( \tan^{-1} a \)
Let \( \tan^{-1} a = \phi \). Then, we have:

\[ \tan \phi = \frac{a}{1} \]

### Step 5: Use angle relationships
Since the sum of angles in a triangle is \( 180^\circ \) and angle \( B \) is \( 90^\circ \):

\[ \theta + \phi = \frac{\pi}{2} \]

Thus, we can express \( \phi \) as:

\[ \phi = \frac{\pi}{2} - \theta \]

### Step 6: Substitute back into the equation
Substituting \( \phi \) and \( \theta \) back into the original equation:

\[ x \left(\frac{\pi}{2} - \theta\right) + y \theta + 2 = 0 \]

### Step 7: Rearrange the equation
Rearranging gives us:

\[ x \frac{\pi}{2} + y \theta - x \theta + 2 = 0 \]

Factoring out \( \theta \):

\[ \theta(-x + y) + x \frac{\pi}{2} + 2 = 0 \]

### Step 8: Determine the concurrency point
For the lines to be concurrent, we need to check specific points given in the options. We will substitute each point into the equation to see if it satisfies it.

#### Checking Point (1/π, 1/π):
Substituting \( x = \frac{1}{\pi} \) and \( y = \frac{1}{\pi} \):

\[ \frac{1}{\pi} \cdot \frac{\pi}{2} + \frac{1}{\pi} \theta - \frac{1}{\pi} \theta + 2 = 0 \]

This simplifies to:

\[ \frac{1}{2} + 2 \neq 0 \]

#### Checking Point (-4/π, -4/π):
Substituting \( x = -\frac{4}{\pi} \) and \( y = -\frac{4}{\pi} \):

\[ -\frac{4}{\pi} \cdot \frac{\pi}{2} - \frac{4}{\pi} \theta + \frac{4}{\pi} \theta + 2 = 0 \]

This simplifies to:

\[ -2 + 2 = 0 \]

This point satisfies the equation.

#### Checking Point (π, π):
Substituting \( x = \pi \) and \( y = \pi \):

\[ \pi \cdot \frac{\pi}{2} + \pi \theta - \pi \theta + 2 \neq 0 \]

### Conclusion
The only point that satisfies the equation is:

\[ \left(-\frac{4}{\pi}, -\frac{4}{\pi}\right) \]

Thus, the correct answer is (b) \((-4/\pi, -4/\pi)\).