# Thread: Need help proving trig identity.

1. ## Need help proving trig identity.

I don't know how to star solving this problem, I'm new in calculus..

f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y

2. What are u trying to find?

3. this one sir..

given: f(x)=tan x ,show that

f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y

4. Whenever I have a trig identity to prove, I usually get everything in terms of sines and cosines. Use the addition of angles formulas. Then simplify.

5. Originally Posted by jasonlewiz
this one sir..

given: f(x)=tan x ,show that

f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y
the result should be $\frac{\sec^2{x} \tan{y}}{1 - \tan{x}\tan{y}}$

6. As Ackbeet said, this is a matter of a trig identity and you can reduce to sine and cosine.

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

Now, divide both numerator and denominator by cos(x)cos(y)
$tan(x+y)= \frac{\frac{sin(x)}{cos(x)}+ \frac{sin(y)}{cos(y)}}{1- \frac{sin(x)}{cos(x)}\frac{sin(y)}{cos(y)}}$
$= \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}$
So that $tan(x+y)- tan(x)= \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- tan(x)$
$= \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- \frac{tan(x)- tan^2(x)tan(y)}{1- tan(x)tan(y)}$

Note that, as skeeter said, you have the sign in the denominator wrong.