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C3 Trig Identity

(Tanx+1)/sinx = (cotx+1)/cosx

I have to prove that the left hand side is identical to the right hand side.

Can someone start me off, i've attempted this about 5 times and just cant get it right!
Multiply the LHS by cosxcosx\frac{\cos x}{\cos x} ?
Reply 2
Original post by Mr M
Multiply the LHS by cosxcosx\frac{\cos x}{\cos x} ?


I did that and got down to (sinx+cosx)/(sinxcosx). I then split it into individual fractions but that did nothing
Original post by ChrisA7x
I did that and got down to (sinx+cosx)/(sinxcosx). I then split it into individual fractions but that did nothing


What must you do to both the numerator and denominator to reach your target?
Original post by ChrisA7x
I did that and got down to (sinx+cosx)/(sinxcosx). I then split it into individual fractions but that did nothing


Well you are there effectively. What is cot x in terms of sin x and cos x?
(edited 12 years ago)
Original post by ChrisA7x
I did that and got down to (sinx+cosx)/(sinxcosx). I then split it into individual fractions but that did nothing


When you've multiplied by that on the LHS you should note that cosx/sinx=cotx.
Reply 6
Original post by ian.slater
What must you do to both the numerator and denominator to reach your target?


Ohh ok divide by sinx/sinx

I just dont think of these when I do the questions, is there any way to know to times by cosx/cosx then divide by sinx/sinx, or is it just trial and error
Original post by ChrisA7x
Ohh ok divide by sinx/sinx

I just dont think of these when I do the questions, is there any way to know to times by cosx/cosx then divide by sinx/sinx, or is it just trial and error


To answer the second bit, I just asked you the question I asked myself.
Original post by ChrisA7x
Ohh ok divide by sinx/sinx

I just dont think of these when I do the questions, is there any way to know to times by cosx/cosx then divide by sinx/sinx, or is it just trial and error


Another way to do it is to simply separate the numerator on the LHS

to get tanx/sinx + 1/sinx

which is 1/cosx + 1/sin x

and you know you must get a cot term. The way of doing that from the above is getting cosx/sinx

you have 1/sinx so multiply all terms by cosx/cosx

See where that gets you :smile:
Original post by ChrisA7x
I just dont think of these when I do the questions, is there any way to know to times by cosx/cosx then divide by sinx/sinx, or is it just trial and error


Just experience. There is nothing wrong with working on both sides of the identity until you get better at it. No A Level examiner will mark you down for this (even though it is considered bad practice).
Original post by Mr M
Just experience. There is nothing wrong with working on both sides of the identity until you get better at it. No A Level examiner will mark you down for this (even though it is considered bad practice).


Are you sure? My maths teacher told me if they say prove an identity is equal to another form on Core 3/4 you must start from the LHS and go to the RHS otherwise you can't get full marks.
Reply 11
Original post by Mr M
Just experience. There is nothing wrong with working on both sides of the identity until you get better at it. No A Level examiner will mark you down for this (even though it is considered bad practice).


ok thanks, I will keep on doing them up until january, will try and memorise some of them
Original post by hassi94
Are you sure? My maths teacher told me if they say prove an identity is equal to another form on Core 3/4 you must start from the LHS and go to the RHS otherwise you can't get full marks.


Yes I'm sure.
Original post by hassi94
Are you sure? My maths teacher told me if they say prove an identity is equal to another form on Core 3/4 you must start from the LHS and go to the RHS otherwise you can't get full marks.


I'm with your Maths teacher. Except that you can start with either side and work towards the other.

In practice, you can play about with both sides in rough. But present the solution properly.
Original post by ian.slater
I'm with your Maths teacher. Except that you can start with either side and work towards the other.

In practice, you can play about with both sides in rough. But present the solution properly.


Well clearly this is desirable but it is not necessary.

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