Answer: see proof below
Step-by-step explanation:
Use the following Double Angle Identities:
sin 2A = 2cos A · sin A
cos 2A = 2 cos²A - 1
Use the following Quotient Identity: tan A = (sin A)/(cos A)
Use the following Pythagorean Identity:
cos²A + sin²A = 1 --> sin²A = 1 - cos²A
Proof LHS → RHS
Given: [tex]\dfrac{1+sin\theta - cos \theta}{1+sin \theta +cos \theta}[/tex]
Let Ф = 2A: [tex]\dfrac{1+sin2A - cos 2A}{1+sin2A +cos2A}[/tex]
Un-factor: [tex]\dfrac{\bigg(\dfrac{1- cos^2\ 2A}{1+cos\ 2A}\bigg)+sin\ 2A }{1+sin\ 2A +cos\ 2A}[/tex]
Pythagorean Identity: [tex]\dfrac{\bigg(\dfrac{sin^2\ 2A}{1+cos\ 2A}\bigg)+sin\ 2A }{1+cos\ 2A +sin\ 2A}[/tex]
Simplify: [tex]\dfrac{sin\ 2A}{1+cos\ 2A}[/tex]
Double Angle Identity: [tex]\dfrac{2sin\ A\cdot cos\ A}{1+(2cos^2 A-1)}[/tex]
Simplify: [tex]\dfrac{2sin\ A\cdot cos\ A}{2cos^2\ A}[/tex]
[tex]=\dfrac{2sin\ A\cdot cos\ A}{2cos^2\ A}[/tex]
[tex]=\dfrac{sin\ A}{cos\ A}[/tex]
Quotient Identity: tan A
[tex]\text{Substitute} A = \dfrac{\theta}{2}}:\qquad tan\dfrac{\theta}{2}[/tex]
[tex]tan\dfrac{\theta}{2} = tan\dfrac{\theta}{2}\quad \checkmark[/tex]
