How do you prove #tan((x/2)+(pi/4))=secx+tanx#?

1 Answer
Sep 14, 2016

Prove trig expression

Explanation:

First, use trig identity:
#tan (a + b) = (tan a + tan b)/(1 - tan a.tan b)#
Since #tan (pi)/4 = 1,# we get
#tan (x/2 + pi/4) = (tan (x/2) + 1)/(1 - tan (x/2)) =#
#= (cos (x/2) + sin (x/2))/(cos (x/2) - sin (x/2)#
Multiply both numerator and denominator by #(cos(x/2) + sin (x/2))#, we get:
Numerator #N = (cos (x/2) + sin (x/2))^2 = 1 + sin x#
Use trig identity: #cos^2 a - sin^2 a = cos 2a#
Denominator #D = cos^2 (x/2) - sin^2 (x/2) = cos x#
#tan (x/2 + pi/4) = N/D = (1 + sin x)/(cos x) = #
#= 1/(cos x) + sin x/(cos x) = sec x + tan x#