Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
tan(x + y) = [tex]\frac{tanx+tany}{1-tanxtany}[/tex] , tan(x - y) = [tex]\frac{tanx-tany}{1+tanxtany}[/tex], tan [tex]\frac{\pi }{4}[/tex] = 1
tan2x = [tex]\frac{2tanx}{1-tan^2x}[/tex]
Consider the left side
tan([tex]\frac{\pi }{4}[/tex] + Θ ) - tan([tex]\frac{\pi }{4}[/tex] - Θ )
= [tex]\frac{tan\frac{\pi }{4}+tan0 }{1-tan\frac{\pi }{4}tan0 }[/tex] - [tex]\frac{tan\frac{\pi }{4}-tan0 }{1+tan\frac{\pi }{4}tan0 }[/tex]
= [tex]\frac{1+tan0}{1-tan0}[/tex] - [tex]\frac{1-tan0}{1+tan0}[/tex]
= [tex]\frac{(1+tan0)^2-(1-tan0)^2}{(1-tan0)(1+tan0)}[/tex]
= [tex]\frac{1+2tan0+tan^20-1+2tan0-tan^20}{1-tan^20}[/tex]
= [tex]\frac{4tan0}{1-tan^20}[/tex]
= 2 [tex]\frac{2tan0}{1-tan^20}[/tex]
= 2 tan2Θ = right side
