Question

Help prove this identitiy

tan(pi/4+x)+tan(pi/4-x)=2sec2x

Answer 1

tan (pi/4+x) = (tan pi/4 + tan x)/(1 -tan pi/4 tan x)
tan (pi/4+x) = (tan pi/4 - tan x)/(1 +tan pi/4 tan x)
note tan pi/4 = 1
add and put over common denominator
[(1+tan x)(1+tan x)+(1-tan x)(1-tan x)]/(1 - tan^2 x)
expand top
[1+2 tanx +tan^2 x +1 - 2 tan x +tan^2 x]/(1 - tan^2 x)
or
2(1 + tan^2 x)/(1-tan^2 x)
2 (1+sin^2 x/cos^2 x)/ (1 - sin^2 x/cos^2 x)
= 2 (cos^2 x + sin^2 x)/(cos^2 x - sin^2 x)
but cos^2 + sin^2 = 1 so
= 2/(cos^2 x - sin^2 x)
Now work on the right
cos 2 x = cos^2 x - sin^2 x
so
sec 2x = 1/(cos^2x-sin^2 x)
2 sec 2x = 2 / (cos^2 x -sin^2 x)
the end

Answer 2

Or

LS = sin(pi/4+x)/cos(pi/4+x) + sin(pi/4-x)/cos(pi/4-x)

I will show the expansion of one of these:
sin(pi/4+x) = sin(pi/4)cosx + cos(pi/4)sinx
= (√2/2)cosx + (√2/2)sinx
there will be a √2/2 in every term of each expansion, so it can be factored out and canceled.
so
LS = (cosx+sinx)/(cosx-sinx) + (cosx-sinx)/(cosx + sinx)
= [(cosx+sinx)^2 + (cosx-sinx)^2]/(cos^2x - sin^2x)
= 2/cos2x = 2sec2x = RS
Q.E.D.

Answer 3

Tan(45+x)+tan(45-x)=2sec2x