If `f(x)=(1+x)/(1-x)`, show that `f|f(tantheta)]=-cottheta.`

Answer

To solve the problem, we need to show that \( f(f(\tan \theta)) = -\cot \theta \) given that \( f(x) = \frac{1+x}{1-x} \).

### Step 1: Find \( f(\tan \theta) \)

We start by substituting \( x = \tan \theta \) into the function \( f(x) \):

\[ f(\tan \theta) = \frac{1 + \tan \theta}{1 - \tan \theta} \]

### Step 2: Substitute \( f(\tan \theta) \) into \( f(x) \)

Next, we need to find \( f(f(\tan \theta)) \). This means we will substitute \( f(\tan \theta) \) into the function \( f(x) \):

Let \( y = f(\tan \theta) = \frac{1 + \tan \theta}{1 - \tan \theta} \).

Now we need to find \( f(y) \):

\[ f(y) = f\left(\frac{1 + \tan \theta}{1 - \tan \theta}\right) = \frac{1 + \frac{1 + \tan \theta}{1 - \tan \theta}}{1 - \frac{1 + \tan \theta}{1 - \tan \theta}} \]

### Step 3: Simplify the numerator

The numerator becomes:

\[ 1 + \frac{1 + \tan \theta}{1 - \tan \theta} = \frac{(1 - \tan \theta) + (1 + \tan \theta)}{1 - \tan \theta} = \frac{2}{1 - \tan \theta} \]

### Step 4: Simplify the denominator

The denominator becomes:

\[ 1 - \frac{1 + \tan \theta}{1 - \tan \theta} = \frac{(1 - \tan \theta) - (1 + \tan \theta)}{1 - \tan \theta} = \frac{-2\tan \theta}{1 - \tan \theta} \]

### Step 5: Combine the results

Now we can combine the results of the numerator and denominator:

\[ f(f(\tan \theta)) = \frac{\frac{2}{1 - \tan \theta}}{\frac{-2\tan \theta}{1 - \tan \theta}} = \frac{2}{-2\tan \theta} = -\frac{1}{\tan \theta} \]

### Step 6: Convert to cotangent

Since \( \frac{1}{\tan \theta} = \cot \theta \), we have:

\[ f(f(\tan \theta)) = -\cot \theta \]

### Conclusion

Thus, we have shown that:

\[ f(f(\tan \theta)) = -\cot \theta \]