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Math Help - trig proof road block

  1. #1
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    trig proof road block

    i have a trig proof that i started but ended up going no where with i got like 4 steps into it and hit a wall. if someone could help me out it would be much appriciated thanx. Sec+2tan = cot+2cos
    >>>>>>>> csc - sin
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by speedy229 View Post
    i have a trig proof that i started but ended up going no where with i got like 4 steps into it and hit a wall. if someone could help me out it would be much appriciated thanx. Sec+2tan = cot+2cos
    >>>>>>>> csc - sin
    I'm kind of confused what you are asking. Are you asking to prove?

    \sec{x} +2\tan{x} = \cot{x} + 2\cos{x}

    If so what relevance does

    \csc{x} - \sin{x}

    have to the problem?
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  3. #3
    o_O
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    I think the OP means: \sec x + 2\tan x = \frac{\cot x + 2\cos x}{\csc x - \sin x}

    Start from the RHS. Convert everything into terms of \sin x and \cos x. Then multiply both top and bottom by \sin x. Split the resulting fraction and simplify.
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  4. #4
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    Hello, speedy229!

    I had to guess at what you meant.

    And be careful . . . \sin has no meaning
    . . . . . just as meaningless as \sqrt{\;\;} or (\;)^2


    \sec x+2\tan x \:=\: \frac{\cot x + 2\cos x}{\csc x - \sin x}

    The right side is: . \frac{\dfrac{\cos x}{\sin x} + 2\cos x}{\dfrac{1}{\sin x} - \sin x}


    Multiply by \frac{\sin x}{\sin x}\!:\quad \frac{\sin x\left(\dfrac{\cos x}{\sin x} + 2\cos x\right)}{\sin x\left(\dfrac{1}{\sin x} - \sin x\right)} \;=\;\frac{\cos x + 2\sin x\cos x}{1-\sin^2\!x} . = \;\frac{\cos x(1 + 2\sin x)}{\cos^2\!x}


    . . . . . = \;\frac{1+2\sin x}{\cos x} \;\;=\;\;\frac{1}{\cos x} + \frac{2\sin x}{\cos x} \;\;=\;\;\sec x + 2\tan x




    Edit: I now see that o_O already gave you the game plan . . .
    .
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