Prove that secB - cosB equals tanBsinB?

Answer 1

Try to write everything in terms of sines and cosines:

1 / cos B - cos B = (sin B / cos B) sin B

1 / cos B - cos B = sin2B / cos B

Multiply by the common denominator, cos B:

1 - cos2B = sin2B

Use the pithagorean identity on the left side:

sin2B + cos2B - cos2B = sin2B

sin2B = sin2B