Simpiifying trig functions...urgent help needed

**aikenfan** · November 24th 2007, 05:34 AM

I have tried these several times and cannot figure out how to simplify these...
There are 2 problems...I would really appreciate it if someone would look them over and guide me...I am having a very hard time with this unit
Thank you

**red_dog** · November 24th 2007, 05:40 AM

$(1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x$

$\displaystyle\cot^2y(\sec^2y-1)=\frac{\cos^2y}{\sin^2y}\left(\frac{1}{\cos^2y}-1\right)=\frac{\cos^2y}{\sin^2y}\cdot\frac{1-\cos^2y}{\cos^2y}=\frac{\sin^2y}{\sin^2y}=1$

**Soroban** · November 24th 2007, 06:01 AM

Hello, aikenfan!

If you're familiar with basic identities, these are simple . . .

(10) Prove: . $\cot^2\!y(\sec^2\!y - 1) \;=\;1$

$\text{We have: }\;\cot^2\!y\underbrace{(\sec^2\!y - 1)}_{\text{This is }\tan^2\!y} \;=\;\cot^2\!y\cdot\tan^2\!y \;=\;(\cot y\cdot\tan y)^2 \;=\;(1)^2 \;=\;1$

**aikenfan** · November 24th 2007, 06:13 AM

Thank you very much for your help...I have been having quite a hard time with these...I know the identities, but I am never sure which way to go with them, and the I end up at a dead end...I appreciate all of your help!

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