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Mar 31 2013 11:42pm
Someone please do these step by step on a piece of paper please im having a hard time!
Thank you and god bless!
http://tinypic.com/r/qzons5/6
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Mar 31 2013 11:43pm
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Mar 31 2013 11:44pm
NEED THIS IN 8 HOURS FROM NOW, IF ITS LATER DON'T BOTHER.
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Apr 1 2013 12:15am
1. This one is tricky....you have to multiply the top and bottom by the following term: 1-cscx

When you do this, you get the following: 1-csc^2 x/[cotx*(1-cscx)]

We know that 1+cot^2 x = csc^2 x....That means 1-csc^2 x is equal to -cot^2 x.

So we have -cot^2 x/[cotx*(1-cscx)]...one cotx term cancels and you get -cotx/(1-cscx)...removing the negative sign from the top and bottom yields cotx/(cscx-1)

We know that cotx = cosx/sinx....so multiply the numerator and denominator by sinx...this yields:

cosx/(cscx*sinx -sinx)....since sinx = 1/cscx...that leaves us with cosx/(1-sinx)


2.

Find the common denominator...which is (1+sinx)cosx

That yields:

cos^2 x / [(1+sinx)cosx] for the first term and (1+2sinx + sin^2 x)/[(1+sinx)cosx] for the second

The numerator then becomes (1+2sinx + sin^2 x) + cos^2 x....since sin^2 x + cos^2 x = 1....this reduces to 1+2sinx+1 which further reduces to 2(1+sinx)

Thus, we have 2(1+sinx)/[(1+sinx)cosx]....the 1+sinx terms cancel out which yields 2/cosx which is also written as 2secx

This post was edited by thundercock on Apr 1 2013 12:21am
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Apr 1 2013 12:32am
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Apr 1 2013 12:45am
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Apr 2 2013 12:08am
Thanks ma dude!
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Apr 3 2013 01:29pm
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Apr 3 2013 03:09pm
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