Prove the identity: SinA CosB+ CosA SinB/SinA+ sinB= sinA- SinB/ sinA cosB – cosA sinB

Question

Prove the identity: SinA CosB+ CosA SinB/SinA+ sinB= sinA- SinB/ sinA cosB – cosA sinB

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Aaliyah 3 days 2021-10-11T09:05:33+00:00 2 Answers 0 views 0

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    2021-10-11T09:06:39+00:00

    Answer:

    Please mark as brainliest answer

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    2021-10-11T09:06:55+00:00

    Answer:

    E = \frac {\cos(A) - \cos(B)}{\sin(A) + \sin(B)}

    Multiplying the numerator and denominator by \cos(A) + \cos(B)

    E = \frac {\cos(A) - \cos(B)}{\sin(A) + \sin(B)} \times \frac {\cos(A) + \cos(B)}{\cos(A) + \cos(B)}

    = \frac {\cos^2(A) - \cos^2(B)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}

    Using the identities \cos^2(A) = 1 - \sin^2(A) and \cos^2(B) = 1 - \sin^2(B)

    E = \frac {1 - \sin^2(A) - 1 + \sin^2(B)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}

    = \frac {\sin^2(B) - \sin^2(A)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}

    = \frac {[\sin(B) - \sin(A)][\sin(B) + \sin(A)]}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}

    Dividing the numerator and denominator by \sin(A) + \sin(B), we have:

    E = \frac {\sin(B) - \sin(A)}{\cos(A) + \cos(B)}

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