Posted by olivia on Friday, March 23, 2012 at 8:08pm.
precal - Reiny, Saturday, March 24, 2012 at 12:45am
LS = cotx cotx
= cot^2 x
This is not an identity the way you typed it
if you meant
cosx/(sinxcotx) = 1
LS = cosx/(sinxcosx/sinx)
See how important it is to use brackets when typing in this way ?
precal - Olivia, Saturday, March 24, 2012 at 1:14pm
I apologize 4 the typo, that's how the work was given to me.
precal - Olivia, Saturday, March 24, 2012 at 1:19pm
But thanks alot!
Answer This Question
More Related Questions
- Precal - Verify the identity: sin^(1/2)x*cosx - sin^(5/2)*cosx = cos^3x sq root ...
- precal - 1-cosx/1+cosx=(cotx-cscx)^2
- precal - 1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-...
- precal - 4(sinx)(cosx)+2(sinx)-2(cosx)-1=0 Find x in radians I can't figure this...
- Pre-Calc - Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= ...
- precal - fixed typo ((sinx + cosx)/(1 + tanx))^2 + ((sinx - cosx)/(1 - cotx))^2...
- Math - Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant ...
- Trigonometry Check - Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [...
- precal - please help i can't find the answer for this question. i can only ...
- Precal/Math - How do i get the left side to equal the right hand side? (Cosx)/(1...