Skip to content
Miscellaneous Exercise 3 (I) · Q80

Q.Select the correct option: If tan⁡A−tan⁡B=x\tan A-\tan B=x and cot⁡B−cot⁡A=y\cot B-\cot A=y then cot⁡(A−B)=…\cot(A-B)=\ldots (A) 1y−1x\dfrac{1}{y}-\dfrac{1}{x} (B) 1x−1y\dfrac{1}{x}-\dfrac{1}{y} (C) 1x+1y\dfrac{1}{x}+\dfrac{1}{y} (D) xyx−y\dfrac{xy}{x-y}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
67% · 80/119 Questions
✓ Free question

Step 1: From cot⁡B−cot⁡A=y\cot B-\cot A=y: 1tan⁡B−1tan⁡A=tan⁡A−tan⁡Btan⁡Atan⁡B=xtan⁡Atan⁡B=y\dfrac{1}{\tan B}-\dfrac1{\tan A}=\dfrac{\tan A-\tan B}{\tan A\tan B}=\dfrac{x}{\tan A\tan B}=y, so tan⁡Atan⁡B=xy\tan A\tan B=\dfrac{x}{y}.

Step 2: cot⁡(A−B)=1+tan⁡Atan⁡Btan⁡A−tan⁡B=1+x/yx=(x+y)/yx=x+yxy=1y+1x\cot(A-B)=\dfrac{1+\tan A\tan B}{\tan A-\tan B}=\dfrac{1+x/y}{x}=\dfrac{(x+y)/y}{x}=\dfrac{x+y}{xy}=\dfrac1y+\dfrac1x.

✓Final answer

Option (C) 1x+1y\dfrac{1}{x}+\dfrac{1}{y}

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.