Math Trivia

Question

0

682

1

Find tan2A when cosA = -3/5 and π/2<A<π.

A) 24/7

B) -7/25

C) 7/24

D) -24/25

Guest Oct 23, 2014

+130511

+5

Best Answer

tanA = -4/3

And using

tan2A = [2tanA ] / [1 - tan ²A] we have

tan2A = [2(-4/3) / [ 1 - (-4/3)²] = [-8/3] / [ 1 - 16/9] = [ -8/3]/[-7/9] = [-8/3]*[-9/7]= 72/21 = 24/9

CPhill Oct 23, 2014

CPhill · Answer 1 · 2014-10-23T10:23:21

tanA = -4/3

And using

tan2A = [2tanA ] / [1 - tan ²A] we have

tan2A = [2(-4/3) / [ 1 - (-4/3)²] = [-8/3] / [ 1 - 16/9] = [ -8/3]/[-7/9] = [-8/3]*[-9/7]= 72/21 = 24/9