If `f(y) = (y)/(sqrt(1-y^(2))), g(y) = (y)/(sqrt(1+y^(2)))`, then (fog) y is equal to

Options

A. `(y)/(sqrt(1-y^(2)))`
B. `(y)/(sqrt(1+y^(2)))`
C. y
D. `((1-y^(2)))/(sqrt(1-y^(2)))`

Explanation

To solve the problem, we need to find \( (f \circ g)(y) \), which means we will substitute \( g(y) \) into \( f(y) \).

### Step-by-step Solution:

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

2. **Substitute \( g(y) \) into \( f(y) \)**:
- We need to find \( f(g(y)) \).
- This means we will replace \( y \) in \( f(y) \) with \( g(y) \):
\[ f(g(y)) = f\left(\frac{y}{\sqrt{1 + y^2}}\right) \]

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

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

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

6. **Simplify the Expression**:
- The square root of \( \frac{1}{1 + y^2} \) is \( \frac{1}{\sqrt{1 + y^2}} \):
\[ f(g(y)) = \frac{\frac{y}{\sqrt{1 + y^2}}}{\frac{1}{\sqrt{1 + y^2}}} = y \]

### Final Result:
Thus, \( (f \circ g)(y) = y \).

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