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)#