Number of solutions of equation `sin(cos^(-1)(tan(sec^(-1)x)))=sqrt(1+x)i s//a r e____`

Explanation

To find the number of solutions for the equation \( \sin(\cos^{-1}(\tan(\sec^{-1}(x)))) = \sqrt{1+x} \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the left-hand side of the equation:
\[ \sin(\cos^{-1}(\tan(\sec^{-1}(x)))). \]
Using the identity \( \sec^{-1}(x) = \tan^{-1}(\sqrt{x^2 - 1}) \), we can express it as:
\[ \tan(\sec^{-1}(x)) = \sqrt{x^2 - 1}. \]
Thus, the equation becomes:
\[ \sin(\cos^{-1}(\sqrt{x^2 - 1})). \]

### Step 2: Simplify further
Next, we can use the identity \( \sin(\cos^{-1}(y)) = \sqrt{1 - y^2} \):
\[ \sin(\cos^{-1}(\sqrt{x^2 - 1})) = \sqrt{1 - (\sqrt{x^2 - 1})^2} = \sqrt{1 - (x^2 - 1)} = \sqrt{2 - x^2}. \]
Now, our equation is:
\[ \sqrt{2 - x^2} = \sqrt{1 + x}. \]

### Step 3: Square both sides
To eliminate the square roots, we square both sides:
\[ 2 - x^2 = 1 + x. \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ x^2 + x - 1 = 0. \]

### Step 5: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2}. \]
This gives us two solutions:
\[ x_1 = \frac{-1 + \sqrt{5}}{2} \quad \text{and} \quad x_2 = \frac{-1 - \sqrt{5}}{2}. \]

### Step 6: Evaluate the solutions
Now we need to check if these solutions fall within the valid ranges for the original functions involved:
- The expression \( \sqrt{2 - x^2} \) is defined for \( 2 - x^2 \geq 0 \) or \( -\sqrt{2} \leq x \leq \sqrt{2} \).
- The expression \( \sqrt{1 + x} \) is defined for \( 1 + x \geq 0 \) or \( x \geq -1 \).

Calculating the approximate values:
- \( x_1 = \frac{-1 + \sqrt{5}}{2} \approx 0.618 \) (valid)
- \( x_2 = \frac{-1 - \sqrt{5}}{2} \approx -1.618 \) (not valid since it is less than -√2)

### Conclusion
Thus, the only valid solution is \( x_1 \). Therefore, the number of solutions to the equation is:
\[ \text{Number of solutions} = 1. \]