The discussion revolves around solving the trigonometric equation ##tanx=\frac{(1+tan1)(1+tan2)-2}{(1-tan1)(1-tan2)-2}##. Participants explored various algebraic manipulations and trigonometric identities, ultimately simplifying the equation to ##tanx=\frac{1-tan3}{1+tan3}##. The solution concluded with the result that ##x=42## degrees, achieved by utilizing the identity ##tan(45^\circ - tan3)##.
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Is that really tan(1) and tan(2), as in 1 and 2 radians?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?
It's in degrees. So i divided both num and den by ##tan1tan2 - 1## and i getSammyS said:Whatever it is, divide the numerator & denominator by ##\ tan1\,tan2 -1 \ ##.
##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.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.