# Solution to this trigonometric equation

• diredragon
In summary, the conversation is about finding the value of x in the equation tanx = ((1+tan1)(1+tan2)-2)/((1-tan1)(1-tan2)-2). The attempt at solving the problem involved multiplying through parentheses and using a trigonometric identity. The simplified form of the equation is (1 + (tan1 + tan2)/(tan1tan2-1))/(1 - (tan1 + tan2)/(tan1tan2-1)). With the help of the tan(a+b) identity, the final value of x is found to be 42 degrees.

## Homework Statement

##tanx=\frac{(1+tan1)(1+tan2)-2}{(1-tan1)(1-tan2)-2}## find x

## Homework Equations

3. The Attempt at a Solution [/B]
I tried multiplying through the paranthesis and arrived at ##tanx=\frac{(tan1tan2-1)+(tan2+tan1)}{(tan1tan2-1)-(tan2+tan1)}## and i don't know if this is any simpler than what was originally set. Is there a trigonometric identity at play here?

diredragon said:

## Homework Statement

##tanx=\frac{(1+tan1)(1+tan2)-2}{(1-tan1)(1-tan2)-2}## find x

## Homework Equations

3. The Attempt at a Solution [/B]
I tried multiplying through the parentheses and arrived at ##tanx=\frac{(tan1tan2-1)+(tan2+tan1)}{(tan1tan2-1)-(tan2+tan1)}## and i don't know if this is any simpler than what was originally set. Is there a trigonometric identity at play here?
Is that really tan(1) and tan(2), as in 1 and 2 radians?

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

SammyS said:
Whatever it is, divide the numerator & denominator by ##\ tan1\,tan2 -1 \ ##.
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?

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

Can you try it again.

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?

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.

diredragon
NascentOxygen said:
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.
##tanx=\frac{1-tan3}{1+tan3}## but that's as far as the simolification goes right? Now the use of calculator is required. It says that the answer is ##x=42## in degrees.

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

NascentOxygen