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    cosecx/cotx+tanx = cosx

    Need help proving this!
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    remember cosec x is 1/sinx

    cotx is cosx/sinx

    tanx is sinx/cosx

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    (Original post by the bear)
    remember cosec x is 1/sinx

    cotx is cosx/sinx

    tanx is sinx/cosx

    Thanks, but i know these already. Please feel free to solve it then, because i can't
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    (Original post by the bear)
    remember cosec x is 1/sinx

    cotx is cosx/sinx

    tanx is sinx/cosx

    Don't worry I've done it, cheers
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    (Original post by frankabagnale)
    Thanks, but i know these already. Please feel free to solve it then, because i can't
    \dfrac{\csc x}{\cot x + \tan x} = \dfrac{1}{\sin x}\cdot \dfrac{1}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}

    Now then... what does the denominator of the second fraction simplify to if you put it under a common denominator?
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    Guys I need help
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    (Original post by RDKGames)
    \dfrac{\csc x}{\cot x + \tan x} = \dfrac{1}{\sin x}\cdot \dfrac{1}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}

    Now then... what does the denominator of the second fraction simplify to if you put it under a common denominator?
    All done, thanks though. I have a much harder one which has the class in confusion, can i send a picture of the question to you? It's difficult to write here as it involves square roots
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    (Original post by frankabagnale)
    All done, thanks though. I have a much harder one which has the class in confusion, can i send a picture of the question to you? It's difficult to write here as it involves square roots
    Go on then, just post a pic here.
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    TIA
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    (Original post by frankabagnale)
    TIA
    \begin{aligned} \dfrac{\tan \theta + \sin \theta}{\cot \theta - \cos \theta} & = \dfrac{\sin \theta}{ \cos \theta} \cdot \dfrac{\frac{1}{\cos \theta}+1}{\frac{1}{\sin \theta} - 1} \\ & = \tan \theta \cdot \dfrac{\sin \theta + \sin \theta \cos \theta }{\cos \theta - \cos \theta \sin \theta} \\ & = \tan \theta \cdot \dfrac{\sin \theta}{\cos \theta} \cdot \dfrac{1+\cos \theta}{1-\sin \theta} \\ & = \tan^2 \theta \cdot \dfrac{1+\cos \theta}{1-\sin \theta} \\ & = \tan^4 \theta \cdot \dfrac{\cos^2 \theta}{\sin^2 \theta} \cdot \dfrac{1+\cos \theta}{1-\sin \theta} \end{aligned}

    Rewrite \cos^2 and \sin^2 in terms of \sin and \cos respectively, employ difference of two squares, simplify.
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    (Original post by yemenia16)
    Guys I need help
    if you are stuck on a maths problem then you can post it on here & we will try to help
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    New proof question. If anyone can do it, please attach a full detailed picture of your answer. TIA
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    (Original post by frankabagnale)
    bump
    Begin by using the difference of two squares twice. Express each bracket in terms of sin and cos in a single fraction. Combine.
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    (Original post by RDKGames)
    Begin by using the difference of two squares twice. Express each bracket in terms of sin and cos in a single fraction. Combine.
    Can you please just show me the full working out? Or at least give me some more detail.
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    (Original post by frankabagnale)
    Can you please just show me the full working out? Or at least give me some more detail.
    Have a go first. I don't see how my hint is insufficient.
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    (Original post by RDKGames)
    Have a go first. I don't see how my hint is insufficient.
    LHS or RHS ?
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    (Original post by frankabagnale)
    LHS or RHS ?
    Start with the LHS.
 
 
 
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