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Question: If A + B + C = 180 then the value of tanA + tanB + tanC is :...

If A + B + C = 180 then the value of tanA + tanB + tanC is :

A

≥ 3√3

B

≥ 2√3

C

3√3

D

2√3

Answer

≥ 3√3

Explanation

Solution

Given A + B + C = 180°. This implies tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C. Assuming A, B, C are acute angles, tan⁡A,tan⁡B,tan⁡C>0\tan A, \tan B, \tan C > 0. By AM-GM inequality: tan⁡A+tan⁡B+tan⁡C3≥tan⁡Atan⁡Btan⁡C3\frac{\tan A + \tan B + \tan C}{3} \ge \sqrt[3]{\tan A \tan B \tan C} Let S=tan⁡A+tan⁡B+tan⁡CS = \tan A + \tan B + \tan C. Then S=tan⁡Atan⁡Btan⁡CS = \tan A \tan B \tan C. S3≥S3\frac{S}{3} \ge \sqrt[3]{S} Cubing both sides: S327≥S  ⟹  S3−27S≥0  ⟹  S(S2−27)≥0\frac{S^3}{27} \ge S \implies S^3 - 27S \ge 0 \implies S(S^2 - 27) \ge 0. Since S>0S > 0, we have S2−27≥0  ⟹  S2≥27  ⟹  S≥27=33S^2 - 27 \ge 0 \implies S^2 \ge 27 \implies S \ge \sqrt{27} = 3\sqrt{3}. Thus, tan⁡A+tan⁡B+tan⁡C≥33\tan A + \tan B + \tan C \ge 3\sqrt{3}.