How do I prove

1/cosB - sinB/cosB x sinB = cosB.

THX.

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

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.