The number of real solutions of `tan^(-1)sqrt(x(x+1))+sin^(-1)sqrt(x^2+x+1)=pi/2` is
`a. zero`
`b`. one
`c`. two
`d`. infinite

Explanation

To solve the equation

\[ \tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2}, \]

we can follow these steps:

### Step 1: Use the identity for sine and cosine inverse
We know that

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}. \]

This implies that if we have

\[ \tan^{-1}(u) + \sin^{-1}(v) = \frac{\pi}{2}, \]

then we can express \(\tan^{-1}(u)\) in terms of \(\cos^{-1}(v)\).

### Step 2: Set \(\theta = \tan^{-1}(\sqrt{x(x+1)})\)
Let

\[ \theta = \tan^{-1}(\sqrt{x(x+1)}). \]

Then, we have

\[ \tan(\theta) = \sqrt{x(x+1)}. \]

### Step 3: Express \(\cos(\theta)\)
From the definition of tangent, we can construct a right triangle where the opposite side is \(\sqrt{x(x+1)}\) and the adjacent side is \(1\). The hypotenuse can be calculated as:

\[ \text{Hypotenuse} = \sqrt{(\sqrt{x(x+1)})^2 + 1^2} = \sqrt{x(x+1) + 1} = \sqrt{x^2 + x + 1}. \]

Thus,

\[ \cos(\theta) = \frac{1}{\sqrt{x^2 + x + 1}}. \]

### Step 4: Substitute back into the equation
Now, substituting back into the original equation gives us:

\[ \cos^{-1}\left(\frac{1}{\sqrt{x^2 + x + 1}}\right) + \sin^{-1}\left(\sqrt{x^2 + x + 1}\right) = \frac{\pi}{2}. \]

### Step 5: Set the two expressions equal
From the identity, we have:

\[ \frac{1}{\sqrt{x^2 + x + 1}} = \sqrt{x^2 + x + 1}. \]

### Step 6: Square both sides
Squaring both sides leads to:

\[ 1 = (x^2 + x + 1). \]

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

\[ x^2 + x + 1 - 1 = 0 \implies x^2 + x = 0. \]

### Step 8: Factor the quadratic
Factoring out \(x\):

\[ x(x + 1) = 0. \]

### Step 9: Solve for \(x\)
This gives us the solutions:

\[ x = 0 \quad \text{or} \quad x = -1. \]

### Conclusion
Thus, the number of real solutions to the equation is **2**.