How do you prove the identity $tan (x - \frac{π}{4}) = \frac{tan x - 1}{1 + tan x}$ $tan (x-pi/4)= (tan x - 1)/(1 + tan x)$ ?

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How do you prove the identity $tan (x - \frac{π}{4}) = \frac{tan x - 1}{1 + tan x}$ $tan (x-pi/4)= (tan x - 1)/(1 + tan x)$ ?

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Answer 1 · 2015-04-16T17:13:48

The formula of difference of the tangent is:

$tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β}$ $tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)$ ,

so, in our case:

$tan (x - \frac{π}{4}) = \frac{tan x - tan (\frac{π}{4})}{1 + tan x tan (\frac{π}{4})} = \frac{tan x - 1}{1 + tan x}$ $tan(x-pi/4)=(tanx-tan(pi/4))/(1+tanxtan(pi/4))=(tanx-1)/(1+tanx)$

and this is because $tan (\frac{π}{4}) = 1$ $tan(pi/4)=1$ .

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How do you prove the identity $tan (x - \frac{π}{4}) = \frac{tan x - 1}{1 + tan x}$ $tan (x-pi/4)= (tan x - 1)/(1 + tan x)$ ?

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