Given `f(x)= sqrt(8/(1-x)+8/(1+x))` and `g(x) = 4/(f(sinx))+4/(f(cosx))` then `g(x)` is

Options

A. `pi/2`
B. `pi`
C. `(3pi)/2`
D. `2pi`

Answer

To solve the problem, we will first simplify the function \( f(x) \) and then use it to find \( g(x) \).

### Step 1: Simplify \( f(x) \)

Given:
\[ f(x) = \sqrt{\frac{8}{1-x} + \frac{8}{1+x}} \]

First, we will take \( 8 \) common and find a common denominator:
\[ f(x) = \sqrt{8 \left( \frac{1}{1-x} + \frac{1}{1+x} \right)} \]

The common denominator of \( 1-x \) and \( 1+x \) is \( (1-x)(1+x) = 1-x^2 \). Thus, we can rewrite the expression:
\[ f(x) = \sqrt{8 \left( \frac{(1+x) + (1-x)}{(1-x)(1+x)} \right)} \]
\[ = \sqrt{8 \left( \frac{2}{1-x^2} \right)} = \sqrt{\frac{16}{1-x^2}} \]

So, we have:
\[ f(x) = \frac{4}{\sqrt{1-x^2}} \]

### Step 2: Find \( f(\sin x) \) and \( f(\cos x) \)

Now, we will calculate \( f(\sin x) \) and \( f(\cos x) \).

1. **Calculate \( f(\sin x) \)**:
\[ f(\sin x) = \frac{4}{\sqrt{1 - \sin^2 x}} = \frac{4}{\sqrt{\cos^2 x}} = \frac{4}{|\cos x|} = \frac{4}{\cos x} \quad \text{(for } \cos x \geq 0\text{)} \]

2. **Calculate \( f(\cos x) \)**:
\[ f(\cos x) = \frac{4}{\sqrt{1 - \cos^2 x}} = \frac{4}{\sqrt{\sin^2 x}} = \frac{4}{|\sin x|} = \frac{4}{\sin x} \quad \text{(for } \sin x \geq 0\text{)} \]

### Step 3: Define \( g(x) \)

Now we can define \( g(x) \):
\[ g(x) = \frac{4}{f(\sin x)} + \frac{4}{f(\cos x)} = \frac{4}{\frac{4}{|\cos x|}} + \frac{4}{\frac{4}{|\sin x|}} \]
\[ = |\cos x| + |\sin x| \]

### Step 4: Analyze the periodicity of \( g(x) \)

The function \( g(x) = |\cos x| + |\sin x| \) is periodic. To find its period, we note that both \( |\sin x| \) and \( |\cos x| \) have a period of \( \pi \). Therefore, \( g(x) \) is also periodic with a period of \( \pi \).

### Conclusion

Thus, the function \( g(x) \) is periodic with a period of \( \pi \).