Question regarding Trig identities (1 Viewer)

[ezzy85]

how would you do this one:

show that:

tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A

thanks

 

[spice girl]

Originally posted by ezzy85
show that:

tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A

use compound angles for both trig expressions on the LHS and see wot we get:

tan(pi/4 + A) = [tan(pi/4) + tan(A)] / [1 - tan(pi/4)tanA]
= (1 + tanA)/(1-tanA)

similarly tan(pi/4 - A) = (1-tanA)/(1+tanA)

subtract one from the other with common denominator:
LHS = [ (1+tanA)^2 + (1-tanA)^2 ]/(1-tan^2(A))
= 4tanA / (1-tan^2(A))

= 2(2tanA)/(1-tan^2(A))
= 2tan2A