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# Maths c3 trig identities watch

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1. show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50

cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)

= -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2x

don't know where to go from there
2. ??
3. (Original post by AndyOC)
show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)= -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2xdon't know where to go from there
I don't understand how you've written the question? Is it cosecx divided by (1+cosecx) - cosecx divided by (1-cosecx)? :s
4. (Original post by Save.Me)
I don't understand how you've written the question? Is it cosecx divided by (1+cosecx) - cosecx divided by (1-cosecx)? :s
yeah that's it, sorry it's not clear
5. (Original post by AndyOC)
show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50

cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)

= -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2x

don't know where to go from there
Make them all into sine and cos e.g. 1-(1/sinx) = (sinx-1)/sinx
6. just replace cosecx with a letter ... w/( 1 + w ) - w/ ( 1 - w ) = 50 and just do normal algebra ?
7. You have caused a lot of confusion. the actual question is AQA MPC3 Jan 2011, Q. 7b

show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50

can be written in the form sec^2x = 25

http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
8. (Original post by kkboyk)
Make them all into sine and cos e.g. 1-(1/sinx) = (sinx-1)/sinx
(Original post by the bear)
just replace cosecx with a letter ... w/( 1 + w ) - w/ ( 1 - w ) = 50 and just do normal algebra ?
Yeah I've done it now anyway, just turned cosec^2 into sin fraction and cot^2 into tan fraction then divided. Thanks for the help

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Updated: October 28, 2015
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