View Full Version : Trigonometric identities
george92
Nov 1, 2008, 10:38 AM
How do I prove
1/cosB - sinB/cosB x sinB = cosB.
THX.
basilrazi
Nov 3, 2008, 01:37 AM
sure here:
1/cosB - sinB/cosB X sinB/1 = cosB/1 (simply fraction)
1-sinB/cosB X sinB/1 = cosB (multiply right hand side only)
[sinB(1-sinB)]/cosB = cosB
[sinB - (sinB)^2]/cosB = cosB (cross multiply)
sinB - (sinB)^2 = (cosB)^2
sinB = (cosB)^2 + (sinB)^2
sinB = 1
B = 90 degrees
ebaines
Nov 3, 2008, 10:38 AM
First, note that sin(B)/cos(B)*sin(B) is sin^2(B)/cos(B)
So you are starting with
1/cos(b) - sin^2(B)/cos(B) = cos(B)
Just multiply through by cos(B) and you should arrive at a very basic trig identity. Hope this helps.