SOLUTION: For the function f(x) = tan x, show that f(x+y)

SOLUTION: For the function f(x) = tan x, show that f(x+y) - f(x)= sec^2 x tan y/1-tan x tan y

Question 1165712: For the function f(x) = tan x, show that f(x+y)
- f(x)= sec^2 x tan y/1-tan x tan y
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
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.
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