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lwilli2 · Mar 17, 2008, 03:50 PM

I'm stuck!

Can someone please help?

I need to verify?

tanx + cotx = 2tanx - cscx secx

and

(csc(squared)x ) (cos(squared)x) - cos(squared)x = -1

Any help is greatly appreciated Thanks

ISneezeFunny · Mar 17, 2008, 03:57 PM

Table of Trigonometric Identities

Hope that helps at all.

galactus · Mar 17, 2008, 04:03 PM

Recheck. Neither of those identities is valid.

#1:

$tan(x)+cot(x)$

$=\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)}$

$=\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}$

$=\frac{1}{sin(x)cos(x)}$

$=sec(x)csc(x)$

#2 is not true. Recheck and make sure you typed it correctly.

lwilli2 · Mar 17, 2008, 04:09 PM

Originally Posted by lwilli2

I'm stuck!

Can someone please help?

I need to verify?

tanx + cotx = 2tanx - cscx secx

and

(csc(squared)x ) (cos(squared)x) - cos(squared)x = -1

Any help is greatly appreciated Thanks

Thanks galactus

I did double check #2. I do not think it is an identity.

They way you worked out #1. My teacher said that we should pick one side and try to manipulate it so it it equals the other side.

so your

secX cscX does not = 2tanX - cscX secX

or am I missing something?

galactus · Mar 17, 2008, 04:17 PM

If the identity were valid, you could plug in something and get the same answer from both

sides. It would appear the 2tan(x) is extraneous. It works out to be just

csc(x)sec(x).

galactus · Mar 18, 2008, 06:30 AM

Here is a valid identity.

$cot^{2}(x)-cot^{2}(x)sec^{2}(x)=-1$

$=\frac{cos^{2}(x)}{sin^{2}(x)}-\frac{\sout{cos^{2}(x)}}{sin^{2}(x)}\cdot\frac{1}{ \sout{cos^{2}(x)}}$

$=\frac{cos^{2}(x)}{sin^{2}(x)}-\frac{1}{sin^{2}(x)}$

$=\frac{\overbrace{cos^{2}(x)-1}^{\text{-sin^2(x)}}}{sin^{2}(x)}$

$=\frac{-sin^{2}(x)}{sin^{2}(x)}=-1$