How do you prove the following trig identity: $tan(x + 45°) - tan(45° - x) ≡ 2tan2x$ ?

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Answer 1 · 2015-04-15T18:32:52

Use $tan(a+b) = (tana+tanb)/(1-tanatanb)$

and $tan(a-b) = (tana-tanb)/(1+tanatanb)$

and $tan (2a) = tan(a+a) = (2tana)/(1-tan^2a)$

And also use: $tan (pi/4) =1$ .

Other than that it's just algebra.

Once you get to

$(tanx + 1)/(1-tanx)-(1-tanx)/(1+tanx)$

you'll want a common denominator, so you get

$((tanx+1)^2-(1-tanx)^2)/(1-tan^2x)$

Now do the algebra to get:

$(4tanx)/(1-tan^2x) = 2( (2tanx)/(1-tan^2x)) = 2 tan2x$

How do you prove the following trig identity: tan(x + 45°) - tan(45° - x) ≡ 2tan2xtan(x + 45°) - tan(45° - x) ≡ 2tan2x?