secx - tan x = tan ( pi/4 - x/2)?

Question

This is a Gr.12 Advanced Functions question?

Answer 1

tan ( pi/4 - x/2)
= [1 - tan(x/2)]/[1 + tan(x/2)]
= [cos(x/2) - sin(x/2)]/[cos(x/2) + sin(x/2)]
= [cos(x/2) - sin(x/2)]^2/cos x
= [1-sin x]/cos x
= sec x - tan x

Answer 2

tan (π/4 - x/2)

Use the tangent addition formula, tan (x+y) = (tan x + tan y)/(1 - tan x tan y)

(tan (π/4) + tan (-x/2))/(1 - tan (π/4) tan (-x/2))

Simplifying:

(1 - tan (x/2))/(1 + tan (x/2))

Multiply both numerator and denominator by cos (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)):

(cos (x/2) - sin (x/2))²/(cos² (x/2) - sin² (x/2))

Simplify using cosine addition formula:

(cos (x/2) - sin (x/2))²/cos x

Expand the numerator:

(cos² (x/2) - 2 cos (x/2) sin (x/2) + sin² (x/2))/cos x

Using the Pythagorean theorem and the sine addition formula:

(1 - 2 cos (x/2) sin (x/2))/cos x
(1 - sin x)/cos x
1/cos x - sin x/cos x
sec x - tan x

And we are done.