## anonymous one year ago Verify the identity. cotx minus pi divided by two. = -tan x

1. anonymous

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

2. inkyvoyd

3. anonymous

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

4. anonymous

cofunction*

5. inkyvoyd

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

6. inkyvoyd

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

7. anonymous

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

8. anonymous

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

9. inkyvoyd

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

10. anonymous

yes

11. inkyvoyd

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

12. Mertsj

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

13. inkyvoyd

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

14. anonymous

lol its no problem and true :p

15. inkyvoyd

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)

16. Mertsj

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

17. Mertsj

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

18. anonymous

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

19. UsukiDoll

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

20. UsukiDoll

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