Skip to content
Miscellaneous Exercise 3 (II) · Q102

Q.Prove the following: tan⁡A+tan⁡(60°+A)+tan⁡(120°+A)=3tan⁡3A\tan A+\tan(60°+A)+\tan(120°+A)=3\tan3A

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
86% · 102/119 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1: Let k=tan⁡3Ak=\tan3A. For any angle θ\theta with tan⁡3θ=k\tan3\theta=k: tan⁡3θ=3t−t31−3t2=k\tan3\theta=\dfrac{3t-t^3}{1-3t^2}=k where t=tan⁡θt=\tan\theta, i.e. t3−3kt2−3t+k=0t^3-3kt^2-3t+k=0.

Step 2: Now 3A, 3(60°+A)=180°+3A, 3(120°+A)=360°+3A3A,\,3(60°+A)=180°+3A,\,3(120°+A)=360°+3A all have the SAME tangent, tan⁡3A=k\tan3A=k (since tangent has period 180°180°).

Step 3: So t=tan⁡A, tan⁡(60°+A), tan⁡(120°+A)t=\tan A,\ \tan(60°+A),\ \tan(120°+A) are the three roots of t3−3kt2−3t+k=0t^3-3kt^2-3t+k=0. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.