Question

tan [π/4+x]/tan[π/4-x]=(1+tan x/1-tan x)^2​

Answer 1

Answer:

Proved

Step-by-step explanation:

Tan( A+B) = tanA + tan B / (1 - tanAtanB)

Tan( A-B) = tan A - tan B /(1+ tanAtanB)

Tan (pi/4 + X ) = tan (45+X)

= tan45 + tanX/(1- tan45tanX)

=(1 +tanX)/(1 - tanX)

Tan( pi/4 - X)= tan(45 - X)

=( tan 45 - tanX)/(1 + tan45tanX)

= (1 - tanX )/(1 + tanX)

LHS = {(1 +tanX)/(1 - tanX)}/{(1 -tanX)/(1 +tanX)}

= {(1 +tanX)/(1 - tanX)}^2 =RHS