Total number of solution of the equation `cos^(-1)((1-x^2)/(1+x^2))=sin^(-1)x` is/are one (b) two (c) three (d) four

Options

A. One
B. Two
C. Three
D. Four

Answer

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

### Step 1: Define the Angle
Let \( a = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \). This implies that:
\[ \cos(a) = \frac{1-x^2}{1+x^2} \]

### Step 2: Construct a Right Triangle
From the definition of cosine, we can construct a right triangle where:
- The adjacent side is \( 1 - x^2 \)
- The hypotenuse is \( 1 + x^2 \)

Using the Pythagorean theorem, we can find the opposite side:
\[ \text{Opposite side} = \sqrt{(1+x^2)^2 - (1-x^2)^2} \]
Calculating this gives:
\[ = \sqrt{(1 + 2x^2 + x^4) - (1 - 2x^2 + x^4)} = \sqrt{4x^2} = 2|x| \]

### Step 3: Find Sine of the Angle
Thus, we can express \( \sin(a) \) as:
\[ \sin(a) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2|x|}{1+x^2} \]

### Step 4: Set Up the Equation
Now, substituting back into the original equation gives:
\[ \sin^{-1}\left(\frac{2|x|}{1+x^2}\right) = \sin^{-1}(x) \]

### Step 5: Eliminate the Inverse Sine
Since both sides are equal, we can equate the arguments:
\[ \frac{2|x|}{1+x^2} = x \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 2|x| = x(1+x^2) \]
This can be split into two cases based on the sign of \( x \).

#### Case 1: \( x \geq 0 \)
Here, \( |x| = x \):
\[ 2x = x(1+x^2) \]
If \( x \neq 0 \), we can divide both sides by \( x \):
\[ 2 = 1 + x^2 \implies x^2 = 1 \implies x = 1 \]
So, we have \( x = 0 \) and \( x = 1 \).

#### Case 2: \( x < 0 \)
Here, \( |x| = -x \):
\[ -2x = x(1+x^2) \]
If \( x \neq 0 \), we can divide both sides by \( x \):
\[ -2 = 1 + x^2 \implies x^2 = -3 \]
This case does not yield any real solutions.

### Step 7: Conclusion
The only solutions we found are \( x = 0 \) and \( x = 1 \). Thus, the total number of solutions to the equation is **2**.

### Final Answer
The total number of solutions of the equation is **two**.

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