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

1 Answer
Nghi N.
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

Related questions
See all questions in Proving Identities
Impact of this question
5093 views around the world
You can reuse this answer
Creative Commons License