# Solution to this trigonometric equation

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1. Apr 29, 2016

### diredragon

1. The problem statement, all variables and given/known data
$tanx=\frac{(1+tan1)(1+tan2)-2}{(1-tan1)(1-tan2)-2}$ find x
2. Relevant equations
3. The attempt at a solution

I tried multiplying through the paranthesis and arrived at $tanx=\frac{(tan1tan2-1)+(tan2+tan1)}{(tan1tan2-1)-(tan2+tan1)}$ and i dont know if this is any simpler than what was originally set. Is there a trigonometric identity at play here?

2. Apr 29, 2016

### SammyS

Staff Emeritus
Is that really tan(1) and tan(2), as in 1 and 2 radians?

3. Apr 29, 2016

### SammyS

Staff Emeritus
Whatever it is, divide the numerator & denominator by $\ tan1\,tan2 -1 \$.

4. Apr 30, 2016

### diredragon

It's in degrees. So i divided both num and den by $tan1tan2 - 1$ and i get
$tanx=\frac{(tan1)^2(tan2)^2-2tan1tan2+1+(tan2)^2tan1-tan2+(tan1)^2tan2-tan1}{(tan1)^(tan2)^2-2tan1tan2+1-(tan2)^2tan1+tan2-(tan1)^2-tan1}$
What should i do now?

5. Apr 30, 2016

### Staff: Mentor

From post #1 and using SammyS's suggestion you should get it into the form: (1 + A)/(1 - A)

Can you try it again.

6. Apr 30, 2016

### diredragon

Oh, well then it is
$tanx=\frac{1+\frac{tan1+tan2}{tan1tan2-1}}{1-\frac{tan1+tan2}{tan1tan2-1}}$
Is there a way to simplify this?

7. Apr 30, 2016

### Staff: Mentor

Ask google to search on tan(a+b) (or some trig expression like that) and see whether you can substitute something for that expression you have for the term I designated by A.

8. Apr 30, 2016

### diredragon

$tanx=\frac{1-tan3}{1+tan3}$ but thats as far as the simolification goes right? Now the use of calculator is required. It says that the answer is $x=42$ in degrees.

9. Apr 30, 2016

### diredragon

Yeah i solved it. I used $tan45-tan3$ and the identity to arrive at $tanx=tan42$