WebNo, you don't necessarily have tan ( A) + tan ( B) + tan ( C) = 0. If a, b, c ∈ R are such that a 3 + b 3 + c 3 = 3 a b c, then either a = b = c or a + b + c = 0. Thus, you can have tan ( A) = tan ( B) = tan ( C), which means A B C is an equilateral triangle. – Batominovski Feb 7, 2024 at 11:42 On the other hand, if tan ( A) + tan ( B) + tan Webtan2A=tan (3A-A)= (tan3A-tanA)/ (1+tan3AtanA) tan2A (1+tan3Atan2A) = tan3A-tanA tan2A + tan2Atan3AtanA = tan3A -tanA tan2Atan3AtanA = tan3A -tanA-tan2A Hence proved Thanks I hope above solution will clear your all doubts. Please feel free to post and ask as much doubts as possible. All the best. Other Related Questions on Trigonometry
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Webtan3A−tanA1 − cot3A−cotA1 =kcot2A, then k is equal to A 4 B 3 C 2 D 1 Medium Solution Verified by Toppr Correct option is D) Given, tan3A−tanA1 − cot3A−cotA1 =kcot2A LHS … WebThe value of tan3A - tan2A - tanA is equal to (A) tan3A tan2A tanA (B) -tan3A tan2A tanA (C) tanA tan2A -tan2A tan3A - tan3A tanA (D) None of these. C banyan tree leaf
SOLUTION: If tan3A/tanA = k, then find cos3A/cosA
WebIf tan3A/tanA = k, then find cos3A/cosA. The numerator and denominator are terms of the double angle formulas: Since there are two formulas we could add and subtract the second terms to make two terms on top and bottom if we first multiplied top and bottom by 2: Now break the multiplication by 2 into the addition of two of them: Reverse the ... Webcos 3A = 1 2 and tan 3A = 1 3A = 60° and 3A = 45° A = 20° and A = 15° Concept: Trigonometric Equation Problem and Solution Is there an error in this question or solution? Chapter 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (C) [Page 298] Q 11.4 Q 11.3 … WebExplanation for the correct option: Step 1: Given that, ∴ tan A 2 = 3 2 We know that, cos 2 θ = 2 cos 2 θ - 1 Step 2: Use this formula cos 2 θ = 1 - 2 sin 2 θ ∴ 1 + cos A 1 - cos A = 2 … psm jouan msc 9