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# A2 C3 Trigonometry Proof watch

1. Hello, TSR!

(1-cosx)/sinx = 1/(cosecx + cotx)

Many thanks!
2. What can you replace cosec(x) and cot(x) with?

Then simplify.
3. Split the LHS into two fractions: 1/sinx - cosx/sinx

cosecx - cotx

multiply by [ (cosecx + cotx) / (cosecx + cotx) ]

[ (cosecx -cotx) . (cosecx + cotx) ] / (cosecx + cotx)

(cosec^2x - cot^2x) / (cosecx + cotx)

1 / (cosecx + cotx)

L.H.S = R.H.S
4. (Original post by Drackore)
Hello, TSR!

(1-cosx)/sinx = 1/(cosecx + cotx)

Many thanks!
There will be a few ways to do this one. I would multiply the top and bottom of the RHS by .
5. (Original post by AmarPatel98)
Split the LHS into two fractions: 1/sinx - cosx/sinx

cosecx - cotx

multiply by [ (cosecx + cotx) / (cosecx + cotx) ]

[ (cosecx -cotx) . (cosecx + cotx) ] / (cosecx + cotx)

(cosec^2x - cot^2x) / (cosecx + cotx)

1 / (cosecx + cotx)

L.H.S = R.H.S
Please don't post full solutions. It's against the rules of this forum.
6. Thanks very much for the replies, guys! Ultimately it was Amar's method that I used, as I'd already made a start with that one but didn't know where to go from from cosecx - cotx
7. (Original post by notnek)
Please don't post full solutions. It's against the rules of this forum.
My apologies... I was unaware of this
8. I'm going to oversimplify; realistically you would do most of this in your head.You could have alternatively split up (1-cosx)/sinx in terms of cosecx & secx then multiply top and bottom by cosecx + cotx .
cosecx + cotx=(1/sinx) +(cosx/sinx) =(1+cosx)/sinx

therefore 1/(cosecx + cotx)= sinx/(1+cosx)
= sinx(1-cosx)/(1+cosx)(1-cosx)
= sinx(1-cosx)/1-(cosx)^2
1-(cosx)^2=1-[1-(sinx)^2)] =(sinx)^2

therefore 1/(cosecx + cotx) =sinx(1-cosx)/(sinx)^2
= (1-cosx)/sinx
L.H.S = R.H.S
9. (Original post by Nick Moore)
I'm going to oversimplify; realistically you would do most of this in your head.You could have alternatively split up (1-cosx)/sinx in terms of cosecx & secx then multiply top and bottom by cosecx + cotx .
cosecx + cotx=(1/sinx) +(cosx/sinx) =(1+cosx)/sinx

therefore 1/(cosecx + cotx)= sinx/(1+cosx)
= sinx(1-cosx)/(1+cosx)(1-cosx)
= sinx(1-cosx)/1-(cosx)^2
1-(cosx)^2=1-[1-(sinx)^2)] =(sinx)^2

therefore 1/(cosecx + cotx) =sinx(1-cosx)/(sinx)^2
= (1-cosx)/sinx
L.H.S = R.H.S
Hey - is there any chance you could provide me with a pointer for this question?

Prove this trig identity:

I started with the left side, and got up to

But not really sure where to go from here to end up with

10. (Original post by Funky_Giraffe)
Hey - is there any chance you could provide me with a pointer for this question?

Prove this trig identity:

I started with the left side, and got up to

But not really sure where to go from here to end up with

SImplifying the line you're up to:

Which I'll now write in another way:

Does that help?
11. (Original post by notnek)
SImplifying the line you're up to:

Which I'll now write in another way:

Does that help?
Thanks very much!

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