Question

verify the identity. justify the steps.

see the equation

Answer 1

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

Answer 2

have you learned the formulas for the sum or difference for cot ?

Answer 3

there is a formula for tan(a-b) (and for cot(a-b)

Answer 4

here is a list
http://www.milefoot.com/math/trig/22anglesumidentities.htm

Answer 5

I think I would first use cot = 1/tan to write
\[ \frac{1}{\tan \left( \theta - \frac{\pi}{2} \right)} = - \tan \theta \]

Answer 6

and because I don't like stuff in the denominator, multiply both sides by
tan( theta - pi/2) to get
\[ - \tan \theta \ \tan \left( \theta - \frac{\pi}{2} \right) = 1 \]

Answer 7

now use the tan of a difference of angles (see link)
\[ \tan(A-B) = \frac{ \tan A - \tan B}{1+\tan A \ \tan B} \]

Answer 8

It occurs to me that this approach runs into trouble because tan(pi/2) is undefined
How about using these identities:
\[ \tan(A) = \frac{1}{\tan\left(\frac{\pi}{2} - A\right) }= \cot\left(\frac{\pi}{2} - A\right)\]
we also use
\[ \cot(-A)= - \cot(A) \]

so
\[ \cot\left(\frac{\pi}{2} - A\right) = \cot\left(-( A- \frac{\pi}{2}) \right) = - \cot\left(A-\frac{\pi}{2} \right)\]

Answer 9

putting that together
\[ \tan(A) = \cot\left(\frac{\pi}{2} - A\right) = - \cot\left(A-\frac{\pi}{2} \right) \]
or rearranging
\[ - \cot\left(A-\frac{\pi}{2} \right) = \tan(A) \\
\cot\left(A-\frac{\pi}{2} \right) = - \tan(A) \]