Question

PLEASE HELP ;'( How Do I prove cot(x-pi/2)=-tanx

Answer 1

\[\cot \left( x -\frac{ \pi }{2}\right)=-tanx\]

Answer 2

How have you defined \(\cot(x)\)?

Answer 3

What identity do you get to assume?

Answer 4

1/sec ?

Answer 5

I define \[
\cot(x) = \tan(\pi/2-x)
\]

Answer 6

which one is that?

Answer 7

Because \(\text {co}\) for any trig function means \(f(\pi/2-x)\).

Answer 8

This is called the co-function identity.

Answer 9

ohhh

Answer 10

So \(\cos(x) = \sin(\pi/2-x)\).

Answer 11

And \(\csc(x) = \sec(\pi/2-x)\)

Answer 12

Also \[
\text{cvs}(x) = \text{ver}(\pi/2-x)
\]

Answer 13

oh ok ^_^ thanks

Answer 14

However, this doesn't fully complete the proof.

Answer 15

y = cot (x - Pi/2) = cos (Pi/2 - x)/sin (x - Pi/2)
Numerator: cos (x - Pi/2) = cos ( Pi/2 - x) = sin x
Denominator: sin (x - Pi/2) = -sin (Pi/2 - x) = -cos x
Finally: y = sin x/(-cos x) = -tan x