$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$ (?)

Question

Prove that, $\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$

I could not do that by trying heart and soul. Please solve it.

Answer 1

We are required to prove that (TPT) $$\frac{\cos 30^{\circ}}{\sin 30^{\circ}} = \frac{\cos 20^{\circ}}{\sin 20^{\circ}} - \frac{1}{\cos 10^{\circ}}$$

i.e., TPT

$$\sin 20^{\circ} \cos 10^{\circ} \cos 30^{\circ} = \sin 30^{\circ} \cos 10^{\circ} \cos 20^{\circ} - \sin 30^{\circ} \sin 20^{\circ}$$

i.e., TPT

$$\sin 30^{\circ} \sin 20^{\circ} = \cos 10^{\circ}(\sin 30^{\circ}\cos 20^{\circ} - \sin 20^{\circ} \cos 30^{\circ})$$

i.e., TPT

$$\sin 30^{\circ} \sin 20^{\circ} = \cos 10^{\circ}\sin 10^{\circ}$$

which is always true since $$\sin 30^{\circ} \sin 20^{\circ} = \frac{1}{2} (2 \sin 10^{\circ} \cos 10^{\circ})$$

Hence the proof.

Answer 2

$$\cot20-\cot30=\dfrac{\sin(30-20)}{\sin20\cdot\sin30}$$

Use $\sin20=\sin(2\cdot10)=?$

and $\cos10=\sin(?)$