# Verifying a Trigonometric Identify: cot(pi/2-x)=tanx I am stuck, I used the difference identity for tan, but tan is undefined at pi/2, so would I just substitute the value for cot pi/2 fir tanA in the tan difference formula? Thanks in advance.

Jul 24, 2018

#### Explanation:

$\cot \left(A - B\right) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$.

Letting $A = \frac{\pi}{2} , B = x$, we have,

$\cot \left(\frac{\pi}{2} - x\right) = \frac{\cot \left(\frac{\pi}{2}\right) \cot x + 1}{\cot x - \cot \left(\frac{\pi}{2}\right)}$.

But, $\cot \left(\frac{\pi}{2}\right) = 0$.

$\therefore \cot \left(\frac{\pi}{2} - x\right) = \frac{0 + 1}{\cot x - 0} = \frac{1}{\cot} x = \tan x$.

Otherwise, $\cot \left(\frac{\pi}{2} - x\right) = \frac{\cos \left(\frac{\pi}{2} - x\right)}{\sin \left(\frac{\pi}{2} - x\right)}$,

$= \frac{\cos \left(\frac{\pi}{2}\right) \cos x + \sin \left(\frac{\pi}{2}\right) \sin x}{\sin \left(\frac{\pi}{2}\right) \cos x - \cos \left(\frac{\pi}{2}\right) \sin x}$,

$= \frac{\left(0\right) \left(\cos x\right) + \left(1\right) \left(\sin x\right)}{\left(1\right) \left(\cos x\right) - \left(0\right) \left(\sin x\right)}$,

$= \sin \frac{x}{\cos} x$,

$= \tan x$,