Skip to content
Exercise 3.1 · Q12

Q.Prove that: tan⁡5A−tan⁡3Atan⁡5A+tan⁡3A=sin⁡2Asin⁡8A\dfrac{\tan5A-\tan3A}{\tan5A+\tan3A}=\dfrac{\sin2A}{\sin8A}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
10% · 12/119 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: tan⁡5A−tan⁡3A=sin⁡5Acos⁡3A−cos⁡5Asin⁡3Acos⁡5Acos⁡3A=sin⁡(5A−3A)cos⁡5Acos⁡3A=sin⁡2Acos⁡5Acos⁡3A\tan5A-\tan3A=\dfrac{\sin5A\cos3A-\cos5A\sin3A}{\cos5A\cos3A}=\dfrac{\sin(5A-3A)}{\cos5A\cos3A}=\dfrac{\sin2A}{\cos5A\cos3A}.

Step 2: tan⁡5A+tan⁡3A=sin⁡5Acos⁡3A+cos⁡5Asin⁡3Acos⁡5Acos⁡3A=sin⁡(5A+3A)cos⁡5Acos⁡3A=sin⁡8Acos⁡5Acos⁡3A\tan5A+\tan3A=\dfrac{\sin5A\cos3A+\cos5A\sin3A}{\cos5A\cos3A}=\dfrac{\sin(5A+3A)}{\cos5A\cos3A}=\dfrac{\sin8A}{\cos5A\cos3A}. …

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.