How do you verify the identity #tan(x+45)=(1+tanx)/(1-tanx)#?

2 Answers
Dec 17, 2016

We know that #tantheta = sintheta/costheta#, so:

#sin(x + 45)/cos(x + 45) = (1 + sinx/cosx)/(1 - sinx/cosx)#

We use the sum formulae #sin(A + B) = sinAcosB + cosAsinB# and #cos(A + B) = cosAcosB - sinAsinB# to expand.

#(sinxcos(45) + cosxsin(45))/(cosxcos(45) - sinxsin(45)) = ((cosx + sinx)/cosx)/((cosx - sinx)/cosx)#

#(sinx(1/sqrt(2)) + cosx(1/sqrt(2)))/(cosx(1/sqrt(2)) - sinx(1/sqrt(2))) = (cosx + sinx)/(cosx) xx (cosx)/(cosx - sinx)#

#(1/sqrt2(sinx + cosx))/(1/sqrt(2)(cosx - sinx)) = (cosx + sinx)/(cosx - sinx)#

#(sinx + cosx)/(cosx- sinx) = (cosx+ sinx)/(cosx - sinx)#

#LHS = RHS#

Identity proved!

Hopefully this helps!

Dec 17, 2016

Prove trig expression

Explanation:

Use the trig identity:
#tan (a + b) = (tan a + tan b)/(1 - tan a.tan b#
We get:
#tan (x + 45) = (tan 45 + tan x)/(1 - tan x.tan 45)#
Trig table --> tan 45 = 1
There for:
#tan (x + 45) = (1 + tan x)/(1 - tan x)#