Math, asked by s4us9trupalmuruml, 1 year ago

CosA/1-tanA - sin2A/cosA - sinA = sinA+cosA.

Answers

Answered by ARoy

165

cosA/1-tanA-sin²A/cosA-sinA
=cosA/{1-(sinA/cosA)}-sin²A/(cosA-sinA)
=cosA/{(cosA-sinA)/cosA}-sin²A/(cosA-sinA)
=cos²A/(cosA-sinA)-sin²A/(cosA-sinA)
=(cos²A-sin²A)/(cosA-sinA)
=(cosA+sinA)(cosA-sinA)/(cosA-sinA)
=sinA+cosA (Proved)

Answered by reachemmanuelm

cosA/1-tanA-sin²A/cosA-sinA

=cosA/{1-(sinA/cosA)}-sin²A/(cosA-sinA)

=cosA/{(cosA-sinA)/cosA}-sin²A/(cosA-sinA)

=cos²A/(cosA-sinA)-sin²A/(cosA-sinA)

=(cos²A-sin²A)/(cosA-sinA)

=(cosA+sinA)(cosA-sinA)/(cosA-sinA)

=sinA+cosA (Proved)

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