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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.