If `f: R to R` is defined by `f(x)=(1-x^(2))/(1+x^(2))` then show that `f(tantheta)=cos2theta.`

Answer

To show that \( f(\tan \theta) = \cos(2\theta) \) for the function \( f(x) = \frac{1 - x^2}{1 + x^2} \), we can follow these steps:

### Step 1: Write down the function
We are given the function:
\[ f(x) = \frac{1 - x^2}{1 + x^2} \]

### Step 2: Substitute \( x \) with \( \tan \theta \)
Now, we need to evaluate \( f(\tan \theta) \):
\[ f(\tan \theta) = \frac{1 - (\tan \theta)^2}{1 + (\tan \theta)^2} \]

### Step 3: Use the identity for \( \cos(2\theta) \)
We know from trigonometric identities that:
\[ \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
This matches the form we have in \( f(\tan \theta) \).

### Step 4: Compare the two expressions
Now, we can see that:
\[ f(\tan \theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos(2\theta) \]

### Conclusion
Thus, we have shown that:
\[ f(\tan \theta) = \cos(2\theta) \]