Question

I need help... please help me through this trigonometry. I need full answer.​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

tan(x + y) = $\frac{tanx+tany}{1-tanxtany}$ , tan(x - y) = $\frac{tanx-tany}{1+tanxtany}$, tan $\frac{\pi }{4}$ = 1

tan2x = $\frac{2tanx}{1-tan^2x}$

Consider the left side

tan($\frac{\pi }{4}$ + Θ ) - tan($\frac{\pi }{4}$ - Θ )

= $\frac{tan\frac{\pi }{4}+tan0 }{1-tan\frac{\pi }{4}tan0 }$ - $\frac{tan\frac{\pi }{4}-tan0 }{1+tan\frac{\pi }{4}tan0 }$

= $\frac{1+tan0}{1-tan0}$ - $\frac{1-tan0}{1+tan0}$

= $\frac{(1+tan0)^2-(1-tan0)^2}{(1-tan0)(1+tan0)}$

= $\frac{1+2tan0+tan^20-1+2tan0-tan^20}{1-tan^20}$

= $\frac{4tan0}{1-tan^20}$

= 2 $\frac{2tan0}{1-tan^20}$

= 2 tan2Θ = right side