If you have a typo in #1 and you meant:
cot^2y(sec^2 y-1)=1 , then
LS = cot^2 y (tan^2 y)
= 1
= RS
recall that :
tanx = sinx/cosx and cotx = cosx/sinx
so tanxcotx = (sinx/cosx)(cosx/sinx) = 1
The first substitution is one of the basic Pythagorean identities derived from
sin^2 x + cos^2 x = 1, dividing each term by cos^2 x
#2. Again, I think you have a typo (no brackets)
and it could be
cosx - cosx/(1-tanx) = ...
or on the right side: sinxcosx/(sinx - cosx)
fix the equation with proper brackets and I will attempt it
Hi! Okay so we just learned this last week and I completely lost on what to do. These are some warm up problems we did in class but I still don't know how to do them? Can someone help me? Thanks! :)
Directions: Verify the idenity for each of the problems. Show all work!
1.)cot^2y(sec^2-y-1)=1
2.) cosx-cosx/1-tanx=sin xcoxc/sinx-cosx
3 answers
Oh, yes, I apologize for not including those brackets! I wrote that wrong. The right way is:
cosx - cosx/(1-tanx) = sinxcosx/(sinx - cosx)
cosx - cosx/(1-tanx) = sinxcosx/(sinx - cosx)
cosx - cosx/(1-tanx) = sinxcosx/(sinx - cosx)
LS = cosx - cosx/(1 - sinx/cosx)
= cosx - cosx/( (cosx - sinx)/cosx )
= cosx - cosx(cosx)/(cosx - sinx)
= cosx + cos^2 x/(sinx - cosx)
= (sinxcosx - cos^2 x + cos^2 x)/(sinx - cosx)
= sinxcosx/(sinx - cosx)
= RS
LS = cosx - cosx/(1 - sinx/cosx)
= cosx - cosx/( (cosx - sinx)/cosx )
= cosx - cosx(cosx)/(cosx - sinx)
= cosx + cos^2 x/(sinx - cosx)
= (sinxcosx - cos^2 x + cos^2 x)/(sinx - cosx)
= sinxcosx/(sinx - cosx)
= RS