So I just need confirmation that I'm doing this pre-calc right:

Verify the identity.

cotx minus pi divided by two. = -tan x

Question

So I just need confirmation that I'm doing this pre-calc right:

Verify the identity.

cotx minus pi divided by two. = -tan x

anonymous · Answer

$$\csc (x-\pi/2)=-tanx$$

anonymous · Answer

sorry cot not csc

anonymous · Answer

So I think the proper way to start it is with the even/odd identity of cot(-x)=-cotx but im not sure if the proper way to state this would be -cotxpi/2=-tanx or -cot(x+pi/2)

anonymous · Answer

The first thing I would do is simplify everything into terms of sin and cos
1/cot(x+pi/2) = -sinx/cos x

anonymous · Answer

Rather, cos(x+pi/2)/sin(x+pi/2)

anonymous · Answer

Wait, you've switched from a minus in the initial posting to a plus. Which is it?

anonymous · Answer

well it is originally a - but my teacher said to start with even/odd identities and im trying to figure out if that would automatically make it a positive

anonymous · Answer

For right now, I would leave it as stated, unless you foresee that being the first step in solving the problem.

xapproachesinfinity · Answer

cotx is odd so cot(-x)=-cotx

anonymous · Answer

but what about the pi/2?

xapproachesinfinity · Answer

cot(x-pi/2) is this the initial LHD

anonymous · Answer

yes

zepdrix · Answer

Tangent and Cotangent are co-functions.
$\Large\rm \cot(90-x)=\tan x$

This is almost something that might help you :)

xapproachesinfinity · Answer

you don't need odd even identities
you need cofunction identities or whatever they are called

anonymous · Answer

yeah but the part inside the parenthesis is backward from the confunction identity listed above

xapproachesinfinity · Answer

yeah like what @zepdrix  said

xapproachesinfinity · Answer

oh my bad you need those odd even identities
cot(x-pi/2)=-cot(pi/2-x)

xapproachesinfinity · Answer

so what's next

anonymous · Answer

divide by negative one would b the most logical, right?

anonymous · Answer

both sides of course

xapproachesinfinity · Answer

well look at what @zepdrix  wrote above? doesn't that look the same

anonymous · Answer

Well in the rewritten side, what you wrote had a - in front of it and the identity does not. and so does tanx on the RHS so it would work

xapproachesinfinity · Answer

well you are trying to get -tanx
we have cot(pi/2-x)=tanx so??

xapproachesinfinity · Answer

that negative is needed no?

anonymous · Answer

Yes I think so. Thanks guys!

xapproachesinfinity · Answer

good catch! you are welcome^_^