$\displaystyle tan(x-\frac{\pi}{2})= - cot x$
$\displaystyle \frac{tanx-tan\frac{\pi}{2}}{1+tanx+tan\frac{\pi}{2}}= -cotx$
$\displaystyle \frac{1}{1+0}$
I cant solve this identity can someone help me?
Hello, purplec16!
You can't use that compound-angle formula
. . because $\displaystyle \tan\tfrac{\pi}{2}$ is undefined.
$\displaystyle \tan\left(x-\tfrac{\pi}{2}\right) \:=\: -\cot x$
$\displaystyle \tan\left(x-\tfrac{\pi}{2}\right) \;=\;\frac{\sin(x-\frac{\pi}{2})}{\cos(x-\frac{\pi}{2})} \;=\;\frac{\sin x\cos\frac{\pi}{2} - \cos x\sin\frac{\pi}{2}}{\cos x\cos\frac{\pi}{2} + \sin x\sin\frac{\pi}{2}} $ .$\displaystyle =\; \frac{\sin x\cdot 0 - \cos x\cdot 1}{\cos x\cdot 0 + \sin x\cdot 1} \;=\;-\frac{\cos x}{\sin x} \;=\;-\cot x $
TKH is NOT implying that at all!!
For goddness sake, please read and try to process the help that people are taking the time to give you.
Look at the timestamps on the replies by Soroban and TKH .... Soroban has actually given you the solution. And while he was giving it to you, TKH was typing a hint for you to think about.
The only thing that defies understanding here is why you are confused.