Skip to content
EXERCISE 9.1 · Q7

Q.sec⁡(5x−2)\sec(5x-2)

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
9% · 7/75 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 →

Let f(x)=sec⁡(5x−2)=1cos⁡(5x−2)f(x)=\sec(5x-2)=\dfrac1{\cos(5x-2)}, so f(x+h)=1cos⁡(5x+5h−2)f(x+h)=\dfrac1{\cos(5x+5h-2)}.

f(x+h)−f(x)=cos⁡(5x−2)−cos⁡(5x+5h−2)cos⁡(5x+5h−2)cos⁡(5x−2)=2sin⁡ ⁣(5x+5h2−2)sin⁡ ⁣(5h2)cos⁡(5x+5h−2)cos⁡(5x−2)f(x+h)-f(x)=\dfrac{\cos(5x-2)-\cos(5x+5h-2)}{\cos(5x+5h-2)\cos(5x-2)}=\dfrac{2\sin\!\left(5x+\frac{5h}2-2\right)\sin\!\left(\frac{5h}2\right)}{\cos(5x+5h-2)\cos(5x-2)}

(using cos⁡B−cos⁡A=2sin⁡ ⁣(A+B2)sin⁡ ⁣(A−B2)\cos B-\cos A=2\sin\!\left(\frac{A+B}2\right)\sin\!\left(\frac{A-B}2\right) with A=5x+5h−2, B=5x−2A=5x+5h-2,\ B=5x-2). …

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.