## Trigonometry (11th Edition) Clone

$$\tan\Big(x-\frac{\pi}{2}\Big)=\cot x$$ The statement is false.
$$\tan\Big(x-\frac{\pi}{2}\Big)=\cot x$$ Now what we've already known from the cofunction identities is $$\tan\Big(\frac{\pi}{2}-x\Big)=\cot x$$ Yet here, the statement involves $\tan\Big(x-\frac{\pi}{2}\Big)$, not $\tan\Big(\frac{\pi}{2}-x\Big)$. Unfortunately, 2 expressions are not the same, so we need to find a way to transform $\tan\Big(x-\frac{\pi}{2}\Big)$ to the already known. Though it can be a little bit tricky, we can do like this: $$\tan\Big(x-\frac{\pi}{2}\Big)=\tan\Big[-\Big(\frac{\pi}{2}-x\Big)\Big]$$ And we already know from the negative-angle identities that $$\tan(-\theta)=-\tan\theta$$ Therefore, $$\tan\Big(x-\frac{\pi}{2}\Big)=-\tan\Big(\frac{\pi}{2}-x\Big)$$ $$\tan\Big(x-\frac{\pi}{2}\Big)=-\cot x$$ Since $\cot x\ne-\cot x$, $$\tan\Big(x-\frac{\pi}{2}\Big)\ne\cot x$$ The statement thus is false.