If `f(x)=(x)/(sqrt(1-x^2)) and g(x)=(x)/(sqrt(1+x^2))`, then (fog)(x) ​ =

Options

A. a. x
B. b. 0
C. c. -1
D. d. 2

Answer

To find \( (f \circ g)(x) \), we need to evaluate \( f(g(x)) \).

1. **Identify the functions**:
- \( f(x) = \frac{x}{\sqrt{1 - x^2}} \)
- \( g(x) = \frac{x}{\sqrt{1 + x^2}} \)

2. **Substitute \( g(x) \) into \( f(x) \)**:
- We need to find \( f(g(x)) \):
\[ f(g(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right) \]

3. **Replace \( x \) in \( f(x) \) with \( g(x) \)**:
- Substitute \( g(x) \) into \( f(x) \):
\[ f\left(\frac{x}{\sqrt{1 + x^2}}\right) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 - \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} \]

4. **Simplify the denominator**:
- Calculate \( \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 \):
\[ \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2} \]
- Thus, \( 1 - \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 \) becomes:
\[ 1 - \frac{x^2}{1 + x^2} = \frac{1 + x^2 - x^2}{1 + x^2} = \frac{1}{1 + x^2} \]

5. **Substitute back into the expression**:
- Now substitute this back into our expression for \( f(g(x)) \):
\[ f(g(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{\frac{1}{1 + x^2}}} \]

6. **Simplify further**:
- The square root in the denominator simplifies:
\[ \sqrt{\frac{1}{1 + x^2}} = \frac{1}{\sqrt{1 + x^2}} \]
- Thus, we have:
\[ f(g(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\frac{1}{\sqrt{1 + x^2}}} = x \]

7. **Final result**:
- Therefore, \( (f \circ g)(x) = x \).