Given that (1-cos2a) / (sinacosa) = 1, Tan (B-A) = - 1 / 3, what is Tan (b-2a) equal to?

Given that (1-cos2a) / (sinacosa) = 1, Tan (B-A) = - 1 / 3, what is Tan (b-2a) equal to?

From (1-cos2a) / (sinacosa) = 2Sin ^ 2 (a) / sinacosa = 2tana = 1, there is Tana = 1 / 2
tan( b-2a）=tan( b-a)-tana/1+tan( b-a)tana=(-1/3-1/2)/1-1/3*1/2=-1

Given Sina = 3 / 5, a ∈ (0, Wu / 2), calculate Tan (a + Wu / 4)

From the range of a, cosa = - 4 / 5, Tana = - 3 / 4
From the formula, Tan (a + π / 4) = (Tana + Tan π / 4) / (1-tana &; Tan π / 4)
=(-3/4+1)/(1+3/4)
=1/7

Tan (Wu / 2-A) = 1 / Tan

tan(π/2-A)=cotA=1/tan A

Given Sina cosa = 1 / 5, Wu / 4 ＜ a ＜ Wu / 2, find the value of Tan (Wu / 4 + a). And sin2a = 24 / 25? Is that ok
.

tan(π/4+a)=(1+tana)/(1-tana)=(sina+cosa)/(sina-cosa)=±7
sin2a=2sinacosa=(sina-cosa)²-1=-24/25