Homework Help: Simplifying with Trigonometry Identities

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

Hi, I am a senior in High School having a really difficult time with two problems. I have to prove using the trigonometric identities that they equal each other but I am having a really hard time trying to get them to equal each other. I've spent over 2 hours trying to get these two problems right but I can't. I don't know if the math problems themselves are working because there is some error or that I am just doing them wrong (even though I've spent a long time trying various methods to get them correct). So I was wondering if someone could help me really soon. Here are the problems:
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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