Question 1165712
The equation you provided, $f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}$, is incorrect or requires a different approach to prove. The correct simplification for the left-hand side is generally simpler.

Here is the correct derivation for $f(x+y) - f(x)$ and an explanation of the identity you provided.

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## 1. Correct Derivation for $f(x+y) - f(x)$

Given $f(x) = \tan x$, the expression $f(x+y) - f(x)$ is:

$$f(x+y) - f(x) = \tan(x+y) - \tan x$$

Using the tangent addition formula, $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$:

$$\tan(x+y) - \tan x = \frac{\tan x + \tan y}{1 - \tan x \tan y} - \tan x$$

To combine the terms, find a common denominator:

$$\frac{\tan x + \tan y}{1 - \tan x \tan y} - \frac{\tan x (1 - \tan x \tan y)}{1 - \tan x \tan y}$$

Combine the numerators:

$$\frac{(\tan x + \tan y) - (\tan x - \tan^2 x \tan y)}{1 - \tan x \tan y}$$

Distribute the negative sign in the numerator:

$$\frac{\tan x + \tan y - \tan x + \tan^2 x \tan y}{1 - \tan x \tan y}$$

The $\tan x$ terms cancel out:

$$\frac{\tan y + \tan^2 x \tan y}{1 - \tan x \tan y}$$

Factor out $\tan y$ from the numerator:

$$\frac{\tan y (1 + \tan^2 x)}{1 - \tan x \tan y}$$

Finally, use the Pythagorean identity $\mathbf{1 + \tan^2 x = \sec^2 x}$:

$$f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}$$

## 2. Conclusion

The identity you asked to show, $\mathbf{f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}}$, is **correct**.

Your expression:
$$\mathbf{LHS: f(x+y) - f(x)}$$

Your proposed result:
$$\mathbf{RHS: \frac{\sec^2 x \tan y}{1 - \tan x \tan y}}$$

Since the derivation of $f(x+y) - f(x)$ leads directly to $\frac{\sec^2 x \tan y}{1 - \tan x \tan y}$, the identity is proven.