The number of solutions of the equation `cos^(-1)((1+x^2)/(2x))-cos^(-1)x=pi/2+sin^(-1)x` is 0 (b) 1 (c) 2 (d) 3

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \( \cos^{-1}\left(\frac{1+x^2}{2x}\right) - \cos^{-1}(x) = \frac{\pi}{2} + \sin^{-1}(x) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^{-1}\left(\frac{1+x^2}{2x}\right) - \cos^{-1}(x) = \frac{\pi}{2} + \sin^{-1}(x) \]

### Step 2: Use the identity
We know that:
\[ \cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \]
Thus, we can rewrite the right-hand side:
\[ \frac{\pi}{2} + \sin^{-1}(x) = \cos^{-1}(x) + \sin^{-1}(x) + \sin^{-1}(x) = \cos^{-1}(x) + 2\sin^{-1}(x) \]

### Step 3: Rearranging the equation
Now, we can rearrange our equation:
\[ \cos^{-1}\left(\frac{1+x^2}{2x}\right) = \cos^{-1}(x) + 2\sin^{-1}(x) \]

### Step 4: Simplifying further
Using the identity again, we can express \(2\sin^{-1}(x)\) in terms of cosine:
\[ \cos^{-1}\left(\frac{1+x^2}{2x}\right) = \cos^{-1}(x) + \cos^{-1}(x) = \cos^{-1}(x) + \cos^{-1}(x) = \cos^{-1}(x^2) \]

### Step 5: Set the arguments equal
From the previous step, we have:
\[ \frac{1+x^2}{2x} = -1 \]
This leads to:
\[ 1 + x^2 = -2x \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ x^2 + 2x + 1 = 0 \]

### Step 7: Factor the quadratic
This can be factored as:
\[ (x + 1)^2 = 0 \]

### Step 8: Solve for \(x\)
Thus, we find:
\[ x + 1 = 0 \implies x = -1 \]

### Step 9: Determine the number of solutions
The equation \( (x + 1)^2 = 0 \) has a double root at \( x = -1 \), which means there is only one unique solution.

### Final Answer
The number of solutions of the equation is \( \boxed{1} \).

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