Question

Please Help!!!!
Verify the identity.

cot(x=pi/2)=-tanx

Answer 1

is this \(\cot(x-\frac{\pi}{2})=-\tan(x)\) ?

Answer 2

yes

Answer 3

if you are allowed to use the "cofunction" identity you get it in one step

Answer 4

otherwise, use the "subtraction angle" formula for cotangent

Answer 5

the question did not specify what I can and cannot use to sole the problem

Answer 6

one of the "cofunction" identities gives
\[\cot(\frac{\pi}{2}-x)=\tan(x)\] since cotangent is odd, this tell you
\[\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)=-\tan(x)\]

Answer 7

elsewise you can use
\[\cot(x-\frac{\pi}{2})=\frac{\cos(x-\frac{\pi}{2})}{\sin(x-\frac{\pi}{2})}\]

Answer 8

\[Cot(A-B) = \frac{ cotA.CotB+1 }{ CotB-CotA }\]

Answer 9

then use the "subtraction angle" formula top and bottom
it is a pain to write but it will work

Answer 10

or you can use what @Yahoo! wrote above

Answer 11

could you explain the cofunction identity alittle more to me please?

Answer 12

or you can use a right triangle

Answer 13

from the picture above, it should be clear more or less that \(\tan(x)=\cot(\frac{\pi}{2}-x)\)

Answer 14

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

Answer 15

now since cotangent is odd, and since \(x-\frac{\pi}{2}=-(\frac{\pi}{2}-x)\) you have
\[\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)=-\tan(x)\]

Answer 16

ohh okay thank you so much!!!

Answer 17

cofunction identities are
\[\sin(\frac{\pi}{2}-x)=\cos(x)\] etc

Answer 18

works with any "co"

Answer 19

yw

Answer 20

so as my final answer i would just write

cot(x-pi/2)=-tan x

cot(x-pi/2)= -cos(pi/2-x)=-tan x ?

Answer 21

well we know that

\[\cot(\frac{ \pi }{ 2 }-u)=\tan u\]
So, we know that
\[\cot(x-\frac{ \pi }{ 2 }) = -\tan x\]
:o)