The discussion revolves around the proof of the equation $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$. It includes mathematical reasoning and attempts to derive the relationship using trigonometric identities.
There appears to be agreement on the method used to derive the proof, but the discussion does not clarify whether all participants accept the conclusion as established fact.
The discussion relies on specific trigonometric identities and transformations, and the validity of the proof may depend on the assumptions made regarding the angles involved.
anemone said:Prove that $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$.
kaliprasad said:we have tan 60 - tan 20
=sin 60/cos 60 - sin 20/cos 20
= (sin 60 cos 20 - cos 60 sin 20)/ ( cos 60 cos 20)
= sin (60-20)/ (cos 60 cos 20)
= sin 40/ ( cos 60 cos 20)
= (2 sin 20 cos 20)/( cos 60 cos 20)
= 2 sin 20 / cos 60
= 2 sin 20/ (1/2) = 4 sin 20
hence 4 sin 20 + tan 20 = tan 60 = 3^(1/2)
anemone said:Thanks for participating, kaliprasad and I like your method in cracking this problem so much!