## Algebra and Trigonometry 10th Edition

The identity is verified. $tan(x-\frac{\pi}{2})=-cot~x$
Use: $tan(-u)=-tan~u$ $tan(x-\frac{\pi}{2})=-tan(\frac{\pi}{2}-x)=-cot~x$ Why not use $tan(u-v)=\frac{tan~u-tan~v}{1+tan~u~tan~v}$? $tan(x-\frac{\pi}{2})=\frac{tan~x-tan\frac{\pi}{2}}{1+tan~x·tan\frac{\pi}{2}}$ But, $tan\frac{\pi}{2}$ does not exist.