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Verify that each equation is an identity.

16. 1+tanx/sinx+cosx =secx

ok i have a clue on how to do it. i multiplyed the denominator by sinx-cosx and i also did the top but when i do i get this weird fraction with all these cos and sin and then i get lost...plz help me and explain...

Find a numerical value of one trigonometric function of x.
30. 1+tanx/1+cotx=2
same thing lol..i multiplyed the bottom and top by 1-cotx...then i get stumped...plz explain

  • trig -

    you should use brackets so it looks like
    (1+tanx)/(sinx+cosx) =secx

    you are on the right track, after multiplying top and bottom by sinx - cosx you get

    LS = (1+tanx)(sinx-cosx)/(sin^2 x - cos^2 x)
    = (sinx - cosx + sin^2 x/cosx - sinx)/(sin^2x - cos^2x) after expanding
    = (sin^2x - cos^2)/cosx รท (sin^2x - cos^2x)
    = 1/cosx
    = secx
    = RS

    #30 seems to work the same way.

  • trig -

    i don't understand the second step..did u turn tan into sin/cos..? because im trying to do it and i cant get it

    wat i did for the top is
    and then sinx-cosx+sin^2/cosx-cosx

    can the two cos at the end cancel..thats wats screwing me up i think

  • trig -

    here is my multiplication for the top

    (1+tanx)(sinx-cosx) or
    (1+ sinx/cosx)(sinx-cosx) or
    sinx - cosx + sin^2x/cosx - sinx/cosx * cosx
    = sinx - cosx + sin^2x/cosx - sinx
    = -cosx + sin^2x/cosx , now take a common denominator
    = (-cos^2x + sin^2x)/cosx
    = (sin^2x - cos^2x)/cosx

    now you should be able to follow the rest

  • trig -

    yay thnx!

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