Let `f(x) = tan x, x in (-pi/2,pi/2)and g(x) = sqrt(1-x^2)` then `g(f(x))` is

Options

A. `((sqrt(cos 2x)))/(cosx)`
B. `-((sqrt(cos 2x)))/(cosx)`
C. `((sqrt(cos 2x)))/(|cosx|)`
D. Not defined

Answer

To find \( g(f(x)) \) where \( f(x) = \tan x \) and \( g(x) = \sqrt{1 - x^2} \), we will follow these steps:

### Step 1: Write down the functions
We have:
- \( f(x) = \tan x \)
- \( g(x) = \sqrt{1 - x^2} \)

### Step 2: Substitute \( f(x) \) into \( g(x) \)
We need to find \( g(f(x)) \):
\[ g(f(x)) = g(\tan x) = \sqrt{1 - (\tan x)^2} \]

### Step 3: Simplify \( 1 - (\tan x)^2 \)
Using the identity \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we can rewrite:
\[ 1 - \tan^2 x = 1 - \frac{\sin^2 x}{\cos^2 x} \]
To combine these terms, we find a common denominator:
\[ 1 - \tan^2 x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} \]

### Step 4: Recognize the numerator as a trigonometric identity
The expression \( \cos^2 x - \sin^2 x \) can be rewritten using the double angle identity:
\[ \cos^2 x - \sin^2 x = \cos 2x \]
Thus, we have:
\[ 1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x} \]

### Step 5: Substitute back into \( g(f(x)) \)
Now substituting back into our expression for \( g(f(x)) \):
\[ g(f(x)) = \sqrt{1 - \tan^2 x} = \sqrt{\frac{\cos 2x}{\cos^2 x}} \]

### Step 6: Simplify the square root
We can simplify this further:
\[ g(f(x)) = \frac{\sqrt{\cos 2x}}{\cos x} \]

### Step 7: Determine the sign of \( \cos x \)
Since \( x \) is in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), \( \cos x \) is always positive in this range. Therefore, we can drop the absolute value:
\[ g(f(x)) = \frac{\sqrt{\cos 2x}}{\cos x} \]

### Final Result
Thus, the final result for \( g(f(x)) \) is:
\[ g(f(x)) = \frac{\sqrt{\cos 2x}}{\cos x} \]