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# C3 Trig

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1. So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

Thanks
James

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2. (Original post by james0902)
So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

Thanks
James

Posted from TSR Mobile
Subtract from both sides then it's just when
3. Multiply by sinx/sinx.
Is this question basically proving the identity ?
4. I mean you have to simplify the left to find the right. The two sides cannot interact with one another.

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5. (Original post by B_9710)
Multiply by sinx/sinx.
Is this question basically proving the identity ?
Yes it's proving the identity.

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6. (Original post by james0902)
Yes it's proving the identity.

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Multiply both numerator and denominator of LHS by sinx.
7. (Original post by B_9710)
Multiply both numerator and denominator of LHS by sinx.
I'm really struggling so what would that give me? Would that make

sinx (cotx - cosx)

Is that right?

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8. (Original post by james0902)
I'm really struggling so what would that give me? Would that make

sinx (cotx - cosx)

Is that right?

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Think about what cotx is in terms of cosx and sinx. What is sinxcotx?
9. (Original post by james0902)
Yes it's proving the identity.

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Personally I'd multiply top and bottom by

This gets a nice difference of two square on the denominator which is equal to then the numerator just takes a bit messing around with to get get into a fraction in terms of sine/cosine, by which point loads of things cancel and you're soon left with . But I'm not sure how simple you'd find messing around with a fraction as a numerator, should be straight forward enough if you know what's happening.
10. (Original post by james0902)
So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

Thanks
James

Easy solution!!

1. We start off by multiplying both sides of your expression by (1-Sinx)
2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
4. Hence it simplifies to 1-(Cosx/Cotx).
5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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11. (Original post by BJA98)
Easy solution!!

1. We start off by multiplying both sides of your expression by (1-Sinx)
2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
4. Hence it simplifies to 1-(Cosx/Cotx).
5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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wat

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