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Trig proof

  1. Apr 11, 2013 #1

    trollcast

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Show that:

    $$\tan x\sec^4x\equiv\tan x\sec^2x + \tan^3x\sec^2x$$

    2. Relevant equations

    Trig identities / formulae

    3. The attempt at a solution

    I've got 2 different starts for it but I'm stuck after a few steps with both of them:

    Attempt 1:

    $$\tan x \sec^4 x$$
    $$\frac{\tan x}{\cos^4 x}$$
    $$\frac{\frac{\sin x}{\cos x}}{\cos^4 x}$$
    $$\frac{\sin x \cos^4 x}{\cos x}$$
    $$\sin x \cos^3 x$$

    And then I can't think on anything else for this one.

    Attempt 2:

    $$\tan x \sec^4 x$$
    $$\tan x (\tan^2 x + 1)^2$$
    $$\tan x (\tan^4 x + 2\tan^2 x + 1)$$
    $$\tan^5 x + 2\tan^3 x + \tan x$$

    This one looks a bit closer since its got the higher power tans in it but I can't see where to get the sec terms from?
     
  2. jcsd
  3. Apr 11, 2013 #2

    Curious3141

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    Homework Helper

    Fastest way: split the second term up to ##\displaystyle \tan x\tan^2x\sec^2x##, then use the identity ##\displaystyle \tan^2x = \sec^2x - 1##.
     
  4. Apr 11, 2013 #3

    trollcast

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    Gold Member

    How do you get that for the second term as theres nothing like it in our formula books?
     
  5. Apr 11, 2013 #4

    Mark44

    Staff: Mentor

    If it's not in there, it should be. Certainly you know this one:
    sin2x + cos2x = 1

    If you divide both sides of this equation by cos2x, you get:
    tan2x + 1 = sec2x

    If you divide both sides of the first identity by sin2x, you get:
    1 + cot2x = csc2x

    You should have the first of these memorized. The latter two you can derive quickly.
     
  6. Apr 11, 2013 #5

    Curious3141

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    Homework Helper

    ##\displaystyle \sin^2x + \cos^2x = 1##. Divide throughout by ##\displaystyle \cos^2x##. Rearrange.

    In fact, the tan-sec identify is a well-known one in its own right. So is the cot-cosec one, which you can derive by dividing the above equation by ##\displaystyle \sin^2x## instead of ##\displaystyle \cos^2x##.
     
  7. Apr 12, 2013 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is wrong. It should be
    [tex]\frac{sin(x)}{cos^5(x)}[/tex]

    Have you tried doing the same thing to the right side?
     
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