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# C3 Trig Identities

1. http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF
http://store.aqa.org.uk/qual/gce/pdf...W-MS-JAN11.PDF

7B) Unsure on the last stage of this question as I'm left with:

-2cosec^2x/-cot^2x=50

to get to Sec^2x=25

thanks
2. cosec^2x = 1 + cot^2x

so you have -2(1+cot^2x)/-cot^2x=50

You can get rid of the negatives as they are on both the top and bottom

2((sin^2x+cos^2x)/sin^2x)/cot^2x=50

2(1/sin^2x)/cos^2x/sin^2x=50
The sin^2x cancel out to give you:
2 x1/cos^2x=50
therefore... 2sec^2x=50
sec^2x=25

Hope that makes sense

EDIT: You could also go from
2cosec^2x/cot^2x to 2(1/sin^2x)/cos^2x/sin^2x

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