Help Proving An Identity [A2 Core 3] Watch

Kevlar
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I'm having some issues doing this question with proving an identity it's the first of this type of question i've ever seen and was wondering how i would go about doing it i've had an attempt.

Question



(\mathrm{cosec}^2 \theta-\cot^2 \theta)=\frac{1+\cos^2 \theta}{\sin^2\theta}

My Working



(1+\cot^2 \theta)-(\mathrm{cosec}^2 \theta-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{\tan^2 \theta})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{\sec^2 \theta-1})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{[\frac{1}{\cos^2 \theta}]})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

This is where i get stuck im quite unsure what to do here!
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Totally Tom
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u sure its not cosec^2x+cot^2x?

\displaystyle\,cosec^2{x}+cot^2{  x}=\frac{1}{sin^2{x}}+\frac{cos^  2{x}}{sin^2{x}}=\frac{1+cos^2{x}  }{sin^2{x}}
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wizz_kid
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(Original post by Kevlar)
I'm having some issues doing this question with proving an identity it's the first of this type of question i've ever seen and was wondering how i would go about doing it i've had an attempt.

Question



(\mathrm{cosec}^2 \theta-\cot^2 \theta)=\frac{1+\cos^2 \theta}{\sin^2\theta}

My Working



(1+\cot^2 \theta)-(\mathrm{cosec}^2 \theta-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{\tan^2 \theta})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{\sec^2 \theta-1})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

(1+\frac{1}{[\frac{1}{\cos^2 \theta}]})-(\frac{1}{\sin^2 \theta}-1)=\frac{1+\cos^2 \theta}{\sin^2\theta}

This is where i get stuck im quite unsure what to do here!


there is an easier way :


LHS:

1/sin^2thetha - 1/tan^2 thetha

1/sin^thetha - cos^2thetha/sin^2 thetha


taking sin^2thetha as the common denominator :

(1-cos^2 thetha)/sin^2 thetha



I think on the RHS, it should be (1-cos^2 thetha)/sin&^2thetha
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qgujxj39
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You've got a plus or minus sign wrong somewhere. Divide both the terms in the RHS's numerator by the denominator, and a slightly different version of the identity pops out immediately.
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Kevlar
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oh i was using

 \tan^2 \theta+1=\sec^2 \theta

not

 \cot \theta=\frac{\cos \theta}{\sin \theta}
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wizz_kid
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(Original post by Kevlar)
eh :s im confused lolz



cosec = 1/sin
and
cot = 1/tan

hence using that
LHS:

cosec^2 thetha - cot^2 thetha

1/sin^2thetha - 1/tan^2 thetha ; now u knoe that tan= sin/cos hence 1/tan = cos/sin

1/sin^thetha - cos^2thetha/sin^2 thetha


taking sin^2thetha as the common denominator :

(1-cos^2 thetha)/sin^2 thetha


and as i said, the RHS, it should be (1-cos^2 thetha)/sin&^2thetha
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wizz_kid
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(Original post by Kevlar)
oh i was using

 \tan^2 \theta+1=\sec^2 \theta

not

 \cot \theta=\frac{\cos \theta}{\sin \theta}

thts rite


and cosec= 1/sin
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wizz_kid
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the trick is to look at the 3rd letter :

cosec = 1/sin

sec = 1/cos

cot = 1/tan



and u knoe from c2 that tan= sin/cos

hence 1/tan = 1/sin/cos = cos/sin
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Kevlar
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(Original post by wizz_kid)
cosec = 1/sin
and
cot = 1/tan

hence using that
LHS:

cosec^2 thetha - cot^2 thetha

1/sin^2thetha - 1/tan^2 thetha ; now u knoe that tan= sin/cos hence 1/tan = cos/sin

1/sin^thetha - cos^2thetha/sin^2 thetha


taking sin^2thetha as the common denominator :

(1-cos^2 thetha)/sin^2 thetha


and as i said, the RHS, it should be (1-cos^2 thetha)/sin&^2thetha
I get what your saying but the RHS of the equation is definately correct it is (1+cos^2 theta)/sin^2theta

not 1-cos^2 unless there was a printing error on the question sheet which i highly doubt...
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wizz_kid
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(Original post by Kevlar)
I get what your saying but the RHS of the equation is definately correct it is (1+cos^2 theta)/sin^2theta

not 1-cos^2 unless there was a printing error on the question sheet which i highly doubt...

hmmm what about the LHS? is it cosec^2 thetha + cot^2 thetha ?
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Totally Tom
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look you morons. There is a mistake in the paper it should be a plus.

if it isn't then you get \frac{1-\cos^2{\theta}}{\sin^2{\theta}}=  \frac{\sin^2{\theta}}{\sin^2{\th  eta}}=1
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Kevlar
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(Original post by wizz_kid)
hmmm what about the LHS? is it cosec^2 thetha + cot^2 thetha ?
no its cosec^2 theta - cot^2 theta
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wizz_kid
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(Original post by Totally Tom)
There is a mistake in the paper it should be a plus.
agreed
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wizz_kid
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(Original post by Kevlar)
no its cosec^2 theta - cot^2 theta

it cant be!

coz :

cosec^2 theta - cot^2 theta

= 1 + cot^2 thetha - cot^2 thetha

yea

and the cot's cancel out ..which leaves u with 1
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Kevlar
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(Original post by wizz_kid)
it cant be!

coz :

cosec^2 theta - cot^2 theta

= 1 + cot^2 thetha - cot^2 thetha

yea

and the cot's cancel out ..which leaves u with 1
so can someone show me the working if the question was

(\csc^2 \theta + \cot^2 \theta) = \frac{1+\cos^2 \theta}{\sin^2 \theta}
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wizz_kid
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(Original post by Kevlar)
so can someone show me the working if the question was

(\csc^2 \theta + \cot^2 \theta) = \frac{1+\cos^2 \theta}{\sin^2 \theta}


yea



LHS:

1/sin^2thetha + 1/tan^2 thetha

1/sin^thetha + cos^2thetha/sin^2 thetha


taking sin^2thetha as the common denominator :

(1+cos^2 thetha)/sin^2 thetha
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Kevlar
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Thank you all for your help much appreciated, stupid teachers typed the question wrong lol
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wizz_kid
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(Original post by Kevlar)
Thank you all for your help much appreciated, stupid teachers typed the question wrong lol

no problem as long as u understand whts going on
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Kevlar
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yeah i see whats going on i'll have to look for more types of this question and get some practice x]
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wizz_kid
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(Original post by Kevlar)
yeah i see whats going on i'll have to look for more types of this question and get some practice x]

yea and do some delphis papers ...they are challenging but very helpful
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