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Trig Identities question for test tomorrow Help

  1. Jun 20, 2004 #1
    Trig Identities question for test tomorrow Help!!

    Need some help with these two problems:

    Thanks in advance if you could answer them:

    sin (X) / Cos (x) - 1 = show work

    Sec ^2/ cot (x) - Tan ^3x = Tan X

    show work to prove

    last one

    Sec x - Cos x/tanx = sinx

    show work to prove
  2. jcsd
  3. Jun 20, 2004 #2


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    [tex]\frac{\sec ^2 x}{\cot x} - \tan ^3 x = \tan x[/tex]

    Note, [itex]\cot x[/itex] must not be zero. Now, multiply by [itex]\cot x[/itex]:

    [tex]\sec ^2 x - \tan ^2 x = 1 \ \dots \ (1)[/tex]

    Note, if [itex]\cot x[/itex] were zero, then [itex]\cos x[/itex] would have to be zero (since [itex]\cot x = \frac{\cos x}{\sin x}[/itex]), but since it's not, then [itex]\cos x \neq 0[/itex]. So, we can multiply both sides by [itex]\cos ^2 x[/itex]:

    [tex]1 - \sin ^2 x = \cos ^2 x[/tex]

    [tex]\sin ^2 x + cos ^2 x = 1[/tex]

    This is a basic identity you should know. In fact (1) is a commonly used identity too, but I figured I'd get you down at least this far. I assume you won't have to prove this. If you do, then you know that [itex]\sin x[/itex] is the ratio of the side opposite the angle x in a right triangle to the hypoteneuse. You should also know the definition for [itex]\cos x[/itex]. With these two definitions and the Pythagorean Theorem, you should be able to prove those two identities.

    As a general approach to any of these kinds of problems, express everything in terms of sine and cosine. Mutiplying both sides by [itex]\sin x \cos x[/itex]:

    [tex]\sin x - \cos ^3 x = \sin ^2 x \cos x[/tex]

    [tex]\sin x = \cos x(\cos ^2 x + \sin ^2 x)[/tex]

    [tex]\sin x = \cos x[/tex]

    This is wrong. Try x = 32 degrees. It doesn't work. I guess it's a trick question or you mistyped (or I made a mistake).
  4. Jun 20, 2004 #3
    Yea i'm almost positive that last one doesn't work. If you make that tanx, sinx/cosx you are left with secx = 2sinx, which is not true.
  5. Jun 21, 2004 #4


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    But (sec x- cos x)/tan x= sin x is true.

    As AKG suggested change everything to sin and cos:

    (1/cos x- cos x)/(sin x/cos x)

    = ((1- cos<sup>2</sup> x)/cos x)(cos x/sin x)
    = (sin<sup>2</sup> x/cos x)(cos x/sin x)
    = sin x
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