Question

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

Answer 1

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`