Question

Verify the identity.

cotx minus pi divided by two. = -tan x

Answer 1

\[\cot (x-\frac{ \pi }{ 2 })=-\tan x\]

Answer 2

the answer is eggplants=eggplants

Answer 3

oh come on help me out here :( i don't know how to use the confunction identities for cot.

Answer 4

cofunction*

Answer 5

tan=sin/cos
cot=cos/sin
sin(x-pi/2)=cos(x)
etc

Answer 6

I mean my advice to you is to rewrite cot as cos/sin

Answer 7

cos/sin=cot
so cos=sin or cos = tan or cos = sec
but thats only if cos(pi/2-u) but thats not the case in my problem

Answer 8

my problem is (x-pi/2) not (pi/2-x)

Answer 9

do you know even odd functions? sin(-x)=-sin(x) and cos(-x)=cos(x)

Answer 10

yes

Answer 11

hence sin(x-pi/2)=sin(-(pi/2-x))=-sin(pi/2-x) and apologies for the typo earlier

Answer 12

\[\cot (x-\frac{\pi}{2})=\frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}=\frac{\cos x \cos \frac{\pi}{2}+\sin x \sin \frac{\pi}{2}}{\sin x \cos \frac{\pi}{2}-\cos x \sin \frac{\pi}{2}}\]

Answer 13

o no mertsj typed something latexy that's better than what I have rip

Answer 14

lol its no problem and true :p

Answer 15

really though you need to negate your stuff on the inside and apply even odd functions to tak eout the negative sign and rewrite cot(x-pi/2) to cot(pi/2-x) (expand cot to cos/sin to do this)

Answer 16

\[\frac{\cos x(0)+\sin x(1)}{\sin x(0)-\cos x(1)}=\frac{\sin x}{-\cos x}=-\tan x\]

Answer 17

And now you have proven the given identity once and for all.

Answer 18

ooooh i see i thought you had to make both of them negative thus making it
tan=-tan i get it now
ty!!!

Answer 19

that was an intense one.. It probably needed one of the sum or difference identity formulas. right after rewriting cotx = cosx/sinx

Answer 20

then evaluate it at pi/2 or 90 degrees ... terms cancel and viola -tanx =- tanx