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-   -   Proofs? Identities? (https://www.askmehelpdesk.com/showthread.php?t=143762)

  • Oct 22, 2007, 07:03 PM
    mathretard4592222
    Proofs? Identities?
    Some one teach me precal I can't do any of this.

    Prove the identity:
    (cosx)(tanx+sinxcotx)=sinx+cos^2x (thats cos squared times x)
  • Oct 22, 2007, 07:05 PM
    s_cianci
    Change everything to sin and cos. Recall that tan = sin/cos , cot = cos/sin , sec = 1/cos and csc = 1/sin. Take the left side of the equation, (cosx)(tanx+sinxcotx), do the algebra and it'll fall right into place.
  • Oct 24, 2007, 07:13 AM
    Kirandeep Singh
    (cosx)(tanx + sinxcotx) = sinx + cos^2 x

    Taking left hand side and putting tanx = sinx/cosx and cotx = cosx/sinx

    cosx(sinx/cosx + sinx*cosx/sinx)
    = sinx + cos^2x

    Hence proved
  • Nov 11, 2007, 12:05 PM
    nycfunction
    Quote:

    Originally Posted by mathretard4592222
    Some one teach me precal I can't do any of this.

    Prove the identity:
    (cosx)(tanx+sinxcotx)=sinx+cos^2x (thats cos squared times x)


    tanx = sinx/cosx

    cotx = cosx/sinx

    On the left side:

    (cosx)(tanx+sinxcotx)= cosx(sinx/cosx + sinx(cosx/sinx)

    cosx(sinx/cosx) = sinx... cosx cancels out.

    sinx(cosx/sinx) = cosx(cosx) = cos^2x

    Final answer: sinx + cos^2(x) = right side
  • Nov 11, 2007, 03:49 PM
    terryg752
    (cosx)(tanx+sinxcotx)=sinx+cos^2x

    Left hand side =

    cosx(sinx/cosx + sinx cosx /sinx)

    = sinx + cos^2 x

    = right hand side

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