Sin B ( sin B + cot B ) = 1 + cos A - cos^2 A ?

1 Answer
Hriman
Mar 13, 2018

Sin B ( sin B + cot B ) = 1 + cos B - cos^2 B
sin^2B+cosB/cancel(sinB)*cancel(sinB)= 1 + cos B - cos^2 B
sin^2B+ cosB= 1 + cos B - cos^2 B
Manipulated Pythagorean identity:
1-cos^2x= sin^2x
Substitute 1-cos^2x for sin^2x
(1-cos^2x)+ cosB= 1 + cos B - cos^2 B
1 + cos B - cos^2 B= 1 + cos B - cos^2 B

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