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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