Prove/verify the identities: $tan(x+(3pi)/4) = (tanx-1)/(1+tanx)$ ?

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Answer 1 · 2018-04-06T06:13:35

$LHS=tan(x+(3pi)/4)$

$=(tanx+tan((3pi)/4))/(1-tanxtan((3pi)/4))$

$=(tanx+tan(pi-pi/4))/(1-tanxtan(pi-pi/4))$

$=(tanx-tan(pi/4))/(1-tanx(-tan(pi/4))$

$=(tanx-1)/(1+tanx xx1)$

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

Answer 2 · 2018-04-06T06:18:03

$tan(x+(3pi)/4)$

let $x=A$ and $(3pi)/4 = B$

$=>tan (A+B)$

by the identity,
$color(red)(tan (A+B) = (tanA+ tanB)/(1-tanAtanB)$

therefore,

$tan(x+(3pi)/4) = (tanx+ tan((3pi)/4))/(1-tanxtan((3pi)/4)$

$=> (tanx+ (-1))/(1-tanx(-1))$ $color(white)(dddd$ $["as " color(red)(tan((3pi)/4) = -1)]$

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