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    I need help with proving this identity. Any help would be appreciated.

    

(\mathrm{sin} \theta + \mathrm{cosec} \theta)^2 \equiv \mathrm{sin}^2 \theta + \mathrm{cot}^2 \theta +3

    So far ive done this:

    LHS:
    

\mathrm{sin}^2 \theta + 2\mathrm{sin} \theta \mathrm{cosec} \theta + \mathrm{cosec}^2 \theta





\mathrm{sin}^2 \theta + \frac{2\mathrm{sin} \theta}{\mathrm{sin} \theta} + \frac{1}{\mathrm{sin}^2 \theta}

    But im not sure where to go from here, and help would be appreciated. Thanks.
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    Use cosec^2 \theta = cot^2\theta + 1.
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    (Original post by Elelelel)
    I need help with proving this identity. Any help would be appreciated.

    

(\mathrm{sin} \theta + \mathrm{cosec} \theta)^2 \equiv \mathrm{sin}^2 \theta + \mathrm{cot}^2 \theta +3

    So far ive done this:

    LHS:
    

\mathrm{sin}^2 \theta + 2\mathrm{sin} \theta \mathrm{cosec} \theta + \mathrm{cosec}^2 \theta







\mathrm{sin}^2 \theta + \frac{2\mathrm{sin} \theta}{\mathrm{sin} \theta} + \frac{1}{\mathrm{sin}^2 \theta}

    But im not sure where to go from here, and help would be appreciated. Thanks.

    Rite, you are moving in the right direction.

    Hints : -

    1) What does  \frac {2sin\theta}{sin\theta}  equal?

    2) What does  cosec^2 \theta = in terms of  cot^2 \theta ?
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    And in case you were wondering where the above identity is derived from, it is derived from:

    sin^2\theta + cos^2\theta = 1. Divide through by sin^2\theta and see what happens...
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    You are nearly there.

    Leave the \sin^2\theta term alone as that appears on both sides.

    Get the other two terms over a common denominator of \sin^2\theta.

    1+2\sin^2\theta=(1-\sin^2\theta)+3\sin^2\theta

    I bet you can finish it off now.
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    (Original post by Elelelel)
    I need help with proving this identity. Any help would be appreciated.

    

(\mathrm{sin} \theta + \mathrm{cosec} \theta)^2 \equiv \mathrm{sin}^2 \theta + \mathrm{cot}^2 \theta +3

    So far ive done this:

    LHS:
    

\mathrm{sin}^2 \theta + 2\mathrm{sin} \theta \mathrm{cosec} \theta + \mathrm{cosec}^2 \theta





\mathrm{sin}^2 \theta + \frac{2\mathrm{sin} \theta}{\mathrm{sin} \theta} + \frac{1}{\mathrm{sin}^2 \theta}

    But im not sure where to go from here, and help would be appreciated. Thanks.
    I aplogise now for not using latex as I can't be bothered to try and learn this new language. For simplicities sake I'll say theta = x

    You were at \mathrm{sin}^2 \theta + \frac{2\mathrm{sin} \theta}{\mathrm{sin} \theta} + \frac{1}{\mathrm{sin}^2 \theta}

    The sin squared x is fine.

    Next I used the equation sin squared x + cos squared x = 1 and divided it all by sin squared x.

    This formed 1+cot squared x = 1/sin squared x (as cos/sin = cot)

    so we sub in that for the 1/sin squared x in your last line you got to.

    This forms sin squared x + 3 + cot squared x.

    EDIT: Noticed where I went wrong should work now.
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    (Original post by wizz_kid)
    Rite, you are moving in the right direction.

    Hints : -

    1) What does  \frac {2sin\theta}{sin\theta}  equal?

    2) What does  cosec^2 \theta = in terms of  cot^2 \theta ?
    Your thousandth post! Not a good way to reach it though, seeing as you mispelt 'right' lol.
    • Thread Starter
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    Thanks for your help everyone, managed to work it out.
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    (Original post by Merlinedge)
    I aplogise now for not using latex as I can't be bothered to try and learn this new language. For simplicities sake I'll say theta = x

    You were at \mathrm{sin}^2 \theta + \frac{2\mathrm{sin} \theta}{\mathrm{sin} \theta} + \frac{1}{\mathrm{sin}^2 \theta}

    The sin squared x is fine.

    Next I used the equation sin squared x + cos squared x = 1 and divided it all by sin squared x.

    This formed 1+cot squared x = 1/sin squared x (as cos/sin = cot)

    so we sub in that for the -1/sin squared x in your last line you got to.

    This forms sin squared x + 1 + cot squared x.

    Ah sorrry I don't seem to have got to the answer - not sure if I did anything wrong though, would appreciate if anyone could point it out.
    You've forgotten to add the \frac{2sin\theta}{sin\theta}
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    (Original post by Mathematician!)
    Your thousandth post! Not a good way to reach it though, seeing as you mispelt 'right' lol.
    Thanks lol!

    That was intentional because everytime i try to breakdown a maths question, I can hear my maths teacher say, "riiiiiite" (and takes a deep breath). After a month or 2, i started doing the same lol!
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    (Original post by Elelelel)
    Thanks for your help everyone, managed to work it out.
    No problemo.
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    (Original post by wizz_kid)
    Thanks lol!

    That was intentional because everytime i try to breakdown a maths question, I can hear my maths teacher say, "riiiiiite" (and takes a deep breath). After a month or 2, i started doing the same lol!
    Hehe, but you could have written riiiiight lol :yep:
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    (Original post by Mathematician!)
    Hehe, but you could have written riiiiight lol :yep:

    Right (right? )
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    Another question:

    

\frac{\mathrm{sin} \theta}{1 + \mathrm{cos} \theta} + \frac{1+\mathrm{cos} \theta}{\mathrm{sin} \theta} \equiv 2\mathrm{cosec} \theta

    I have managed to get in down to:

    

\frac{2+2\mathrm{cos} \theta}{\mathrm{sin} \theta(1+\mathrm{cos} \theta)}

    But i dont know how to go any further, any help please.
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    (Original post by Elelelel)
    Another question:

    

\frac{\mathrm{sin} \theta}{1 + \mathrm{cos} \theta} + \frac{1+\mathrm{cos} \theta}{\mathrm{sin} \theta} \equiv 2\mathrm{cosec} \theta

    I have managed to get in down to:

    

\frac{2+2\mathrm{cos} \theta}{\mathrm{sin} \theta(1+\mathrm{cos} \theta)}

    But i dont know how to go any further, any help please.
    You could try cross multiplying the original two fractions in order to get a single fraction.
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    That is what i did and that is what i got, unless i have done something wrong in the working out. but i dont know where to go from there
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    Alternatively, factorise the numerator to get 2(1+cos\theta) and cancel out.
    • Thread Starter
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    Oh yea, thanks for that, worked it out. given rep
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    (Original post by Elelelel)
    Oh yea, thanks for that, worked it out. given rep
    Yes, thank you for the rep, though it is neutral pos lol. The thought is what counts!
 
 
 
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