Question

giving off a medal!
Verify the identity:
cot(x-pi/2)=-tan x

Answer 1

Hint:
\(\huge cot(A-B)=-\frac{1+cot(A)cot(B)}{cot(A)-cot(B)}\)

Answer 2

\[Cot(A-B)=Cos(A-B)/Sin(A-B) \]\[=CosACosB+SinASinB/SinACosB-sinBCosA \]
\[\cos(\pi/2)=0\]
\[\sin(\pi/2)=1\]

Answer 3

Which identity is this?
btw,thnks 4 replying:)

Answer 4

is this a sum and difference formula,@Abmon98 ?

Answer 5

\[Cotx=1/tanx\]
\[tanx=sinx/cosx\]
Addition formula thats subtracting two angles at my level i do not know how to proof it i am sorry. i just learn them.
Sin(A-B)=SinACosB-SinBCosA
Sin(A+B)=SinACosB+SinBCosA
Cos(A-B)=CosACosB+SinASinB
Cos(A+B)=CosACosB-SinASinB

Answer 6

\[Cosx*Cos(\pi/2)+Sinx*Sin(\pi/2)/Sinx*Cos(\pi/2)-Sin(\pi/2)*Cosx\]
\[-Sinx/cosx=-tanx\]

Answer 7

It's Ok,thanks for trying to help me:)

Answer 8

your most welcome :)

Answer 9

The cot(A-B) identity may not be very well-known, but it can be as powerful:
Let B=\(\pi/2\), then cotB=0
\(\large cot(A-B)=-\frac{1+cotA*0}{cotA-0}\)
\(\large =-\frac1{cotA}\)
=-tanA