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Prove that secB - cosB equals tanBsinB?

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Anonymous

∙ 15y ago
Updated: 12/7/2022

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

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∙ 15y ago
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