Question
Prove that:
$\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ=0$

Answer

$\text{L.H.S}=\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ$
$=\tan\Big(\pi+\frac{\pi}{4}\Big)\cot\Big(2\pi+\frac{\pi}{4}\Big)+\tan\Big(4\pi+\frac{\pi}{4}\Big)\cot\Big(4\pi-\frac{\pi}{4}\Big)$
$=\tan\frac{\pi}{4}.\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\times\Big(-\cot\frac{\pi}{4}\Big)$ $\Big(\because\cot\Big(4\pi-\frac{\pi}{4}\Big)=-\cot\frac{\pi}{4}\Big)$
$= 1.1 + 1. (-1)$
$= 0$
$\text{= R.H.S}$
$\text{Proved}$

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