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Question: If the angles A, B, C of a triangle ABC are in arithmetical progression then...

If the angles A, B, C of a triangle ABC are in arithmetical

progression then

A

tan⁡A+tan⁡C−3tan⁡Atan⁡C=3\tan A + \tan C - \sqrt{3}\tan A\tan C = \sqrt{3}

B

tan⁡A+tan⁡C+3tan⁡Atan⁡C=3\tan A + \tan C + \sqrt{3}\tan A\tan C = \sqrt{3}

C

tan⁡A+tan⁡C−3tan⁡Atan⁡C=−3\tan A + \tan C - \sqrt{3}\tan A\tan C = - \sqrt{3}

D

tan⁡A+tan⁡C+3tan⁡Atan⁡C=−3\tan A + \tan C + \sqrt{3}\tan A\tan C = - \sqrt{3}

Answer

tan⁡A+tan⁡C+3tan⁡Atan⁡C=−3\tan A + \tan C + \sqrt{3}\tan A\tan C = - \sqrt{3}

Explanation

Solution

Since A, B, C are in Arithmetical progression

2B=A+C2B = A + C also A + B +C = 1800 so that B = 600

In a triangle ABC, we know

TanA + tanB + tanC = tanAtanBtanC

⇒ tanA+tanC = 3\sqrt{3} (tanAtanC-(1)

⇒ tanA+tanC -3\sqrt{3}tanAtanC = −3- \sqrt{3}