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- Feb 14, 2012

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Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).

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- Thread starter
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- #1

- Feb 14, 2012

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Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).

- Mar 31, 2013

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we have tan 60 - tan 20Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).

=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)

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- Feb 14, 2012

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Thanks for participating,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)

- Mar 31, 2013

- 1,346

Thanks anemone. As long as you ask questions whether I answer or not I shall not get Alzheimer's disease.Thanks for participating,kaliprasadand I like your method in cracking this problem so much!