Skip to content
Exercise 11.5 · Q12

Q.1+cos⁡4xcot⁡x−tan⁡x\dfrac{1+\cos 4x}{\cot x-\tan x}

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
25% · 32/129 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 →

Both the numerator and the denominator simplify with double-angle identities, and the result collapses to a single sine.

Step 1. Simplify the numerator. 1+cos⁡4x=2cos⁡22x1+\cos4x=2\cos^22x.

Step 2. Simplify the denominator. cot⁡x−tan⁡x=cos⁡xsin⁡x−sin⁡xcos⁡x=cos⁡2x−sin⁡2xsin⁡xcos⁡x=cos⁡2x12sin⁡2x=2cos⁡2xsin⁡2x\cot x-\tan x=\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}=\dfrac{\cos2x}{\tfrac12\sin2x}=\dfrac{2\cos2x}{\sin2x}.

Step 3. Divide. 2cos⁡22x2cos⁡2x/sin⁡2x=2cos⁡22x⋅sin⁡2x2cos⁡2x=cos⁡2xsin⁡2x=12sin⁡4x\dfrac{2\cos^22x}{2\cos2x/\sin2x}=2\cos^22x\cdot\dfrac{\sin2x}{2\cos2x}=\cos2x\sin2x=\dfrac12\sin4x. …

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.