How do you prove the identity: (tan(x+y)-tany y)/(1+tan(x+y)tan y) = tan x?

Question

I know "how" to do this, I just can't figure it out. (Same goes for the other one I posted). I know all the identities to use, it's just getting to the end! This is the last one though!

Answer 1

Take the identity tan(x + y) = (tan(x) + tan(y))/(1 - tanx * tan y), valid for x + y different from 2k* pi + pi/2 and different from 2k*pi - pi/2. It follows tan(x - y) = (tan(x) - tan(y))/(1 + tanx * tan y),

Observe that tan(x+y) - tan y = tan(x+y-y) *(1+ tan(x+y) tan(y) ) = tan(x) *(1+ tan(x+y) tan(y) ). Therefore, (tan(x+y) - tan y)/(1+ tan(x+y) tan(y) ) ) = tan x, as desired

Answer 2

Use the known formulae tan = sin/cos, and also sin^2 + cos^2 = 1, replace and simplify.

Answer 3

easy as below

we know Tan(A-B) = Sin (A-B)/cos(A-B)
= (Sin A cos B - cos A sin B)/(Cos A cos B + sin A sin B)
= (tan A-tan B)/(1+Tan A tan B)

put A = X+y and B =y then you get the result

Answer 4

Did you copy the problem correctly? What's tany y? Now your turn to answer.

Answer 5

{tan[x+y]-tany}/1+tan[x+y]tany
={sin(x+y)/cos(x+y)-siny/cosy}/{1+sin[x+y]*siny/cos[x+y]*cosy}
={sin[x+y]*cosy-cos[x+y]*siny}/{cos[x+y]*cosy-sin[-x-y]*siny}
=sin[x+y-y]/cos[y-x-y]
=sinx/cos(-x)
=sinx/cosx
=tanx
qed