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INTEGRATION C4 question. HELP! Watch

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    Hi.
    I have a dilemma.

    In the question above, how does one go from cosec2xtan2x to sec2x? Could someone please provide me with detailed to do so.

    Many Thanks
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    (Original post by greentron6)
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    Hi.
    I have a dilemma.

    In the question above, how does one go from cosec2xtan2x to sec2x? Could someone please provide me with detailed to do so.

    Many Thanks
    \displaystyle \text{cosec}\, x = \frac{1}{\sin x} and \displaystyle \tan x = \frac{\sin x}{\cos x} and \displaystyle \frac{1}{\cos x} = \sec x.

    So \displaystyle \text{cosec} \, x \tan x = \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\cos x} = \cdots

    Can you see now?
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    (Original post by Zacken)
    \displaystyle \text{cosec}\, x = \frac{1}{\sin x} and \displaystyle \tan x = \frac{\sin x}{\cos x} and \displaystyle \frac{1}{\cos x} = \sec x.

    So \displaystyle \text{cosec} \, x \tan x = \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\cos x} = \cdots

    Can you see now?
    Not really, no. Could you explain it with the squared values?
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    (Original post by greentron6)
    Not really, no. Could you explain it with the squared values?
    What...? You square it now...
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    (Original post by Zacken)
    What...? You square it now...

    Sorry.
    Is there another way, in which you could explain this?
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    you could work with the identity 1 + tan2 ≡ sec2
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    (Original post by Zacken)
    \displaystyle \text{cosec}\, x = \frac{1}{\sin x} and \displaystyle \tan x = \frac{\sin x}{\cos x} and \displaystyle \frac{1}{\cos x} = \sec x.

    So \displaystyle \text{cosec} \, x \tan x = \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\cos x} = \cdots

    Can you see now?
    Actually, I understand now! you times both fractions together, cancel out sin2x from top and bottom, leaving 1/cos2x.
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    (Original post by greentron6)

    Sorry.
    Is there another way, in which you could explain this?
    You're giving up too easily Just think about it for a sec. Can you see from my post that \text{cosec} \, x \tan x = \sec x ?

    Now you square both sides of that equation to get (\text{cosec} \, x \tan x)^2 = (\sec x)^2 and that's literally it.

    (BTW, in the future tan2x and tan^2 x are two very different things, please make sure you choose the appropriate formatting)
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    This is so easy. Everyone over-complicated it. LOOK:

    \csc^2{x}\tan^2{x} = \frac{tan^2{x}}{\sin^2{x}} = \frac{1}{\cos^2{x}} = \sec^2{x}
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    (Original post by Tobiq)
    This is so easy. Everyone over-complicated it. LOOK:
    Uh, that's what I did? And the_bear used an equally simple trigonometric identity. Could you elaborate on "over-complicate"?

    \cosec^2{x}\tan^2{x} = \frac{tan^2{x}}{\sin^2{x}} = \frac{1}{\cos^2{x}} = \sec^2{x}
    \cosec isn't a default \LaTeX command...
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    (Original post by Zacken)
    Uh, that's what I did? And the_bear used an equally simple trigonometric identity. Could you elaborate on "over-complicate"?



    \cosec isn't a default \LaTeX command...
    I already changed it, I don't know why you sought the need to try and correct me. I'm not even gonna argue with you; this is a very simple problem, that I coherently went through, in one line.
    Also, the bear simply stated a trig identity, that frankly doesn't need to be used, as it goes a more complicated route.
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    (Original post by Tobiq)
    I already changed it, I don't know why you sought the need to try and correct me. I'm not even gonna argue with you; this is a very simple problem, that I coherently went through, in one line.
    Sure. You posted the same method as an existing answer except with less detail and called the existing answers 'over-complicated'.
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    (Original post by Zacken)
    Just think about it for a sec.
    I see what you did there.
 
 
 
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