Prove that $\cot{70°}+4\cos{70°}= \sqrt{3}$

Question

I was able to do only this. Please help with an explanation. This is a homework question.

Answer 1

$$LHS=\cot 70^0+4\cos 70^0$$

$$=\frac{\cos 70^0+4\cos 70^0\sin 70^0}{\sin 70^0}$$

$$=\frac{\cos 70^0+2\sin 140^0}{\sin 70^0}$$

$$=\frac{\cos 70^0+2\sin 40^0}{\sin 70^0}$$

$$=\frac{\cos 70^0+2\cos 50^0}{\sin 70^0}$$

$$=\frac{\cos 70^0+\cos 50^0+\cos 50^0}{\sin 70^0}$$

$$=\frac{2\cos 60^0\cos 10^0+\cos 50^0}{\sin 70^0}$$

$$=\frac{\cos 10^0+\cos 50^0}{\sin 70^0}$$

$$=\frac{2\cos 30^0\cos 20^0}{\sin 70^0}$$

$$=\frac{2\cos 30^0\sin 70^0}{\sin 70^0}$$

$$={2\cos 30^0}$$

$$=\sqrt{3}$$

Answer 2

$\sqrt {3}=\dfrac {\cos 30} {\sin 30}$

$\dfrac {\cos 70} {\sin 70}-\dfrac {\cos 30} {\sin 30}+4\cos 70$=0

$\dfrac {\cos 70\sin 30-cos30\sin 70} {sin 30\sin 70}$+$4\cos 70$=0

$\sin 40+4\cos 70\sin 70 \sin 30$=0

$\sin 40+2\sin 140\sin 30$=0

$\sin 40-2\cos 50\dfrac {1} {2}$=0

sin40=cos50 $\square$