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December 20, 2014

December 20, 2014

Posted by **amanda** on Wednesday, March 7, 2007 at 6:33pm.

Thank you so much!

1: (1+Cotx)^2/Tanx = CosxCsc^3x + 2Cot^2x

2: Cosx/1-Sinx - Tanx = Secx

#2:

Here are some basic trig identities to help with this problem.

sin^2x + cos^2x = 1

secx = 1/cosx

tanx = sinx/cosx

Let's try to get the left side to look like the right side.

cosx/(1-sinx) - tanx =

cosx/(1-sinx) - sinx/cosx =

Common denominator is: (1-sinx)(cosx)

Therefore:

cos^2x/(1-sinx)(cosx) - sinx(1-sinx)/(1-sinx)(cosx) =

[cos^2x - (sinx - sin^2x)]/(1-sinx)(cosx) =

(cos^2x + sin^2x - sinx)/(1-sinx)(cosx) =

(1 - sinx)/(1-sinx)(cosx) =

1/cosx =

secx = secx

I hope this helps with this one; perhaps someone else can help with the first problem.

- Simplifying with Trigonometry Identities -
**Anonymous**, Tuesday, May 31, 2011 at 5:02pm(1+cotx)^2/tanx

=(1+2cotx+cot^2x)/(sinx/cosx)

=[(1+cot^2x)+2cotx]/(sinx/cosx)

=csc^2x/(sinx/cosx)+2cotx/(sinx/cosx)

=csc^2xcosx/(1/sinx)+2cotxcosx/sinx

=csc63xcosx+2cot^2x

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