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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

If A- B=x and B- A=y , then what is the value of (A-B) ?

If tan⁡A−tan⁡B=x\tan \mathrm{A}-\tan \mathrm{B}=x and cot⁡B−cot⁡A=y\cot \mathrm{B}-\cot \mathrm{A}=y, then what is the value of cot⁡(A−B)\cot (A-B) ?

Options

  1. A1x+1y\frac{1}{x}+\frac{1}{y}
  2. B1y−1x\frac{1}{y}-\frac{1}{x}
  3. Cxyx+y\frac{x y}{x+y}
  4. D1+1xy1+\frac{1}{x y}

Correct answer

A. 1x+1y\frac{1}{x}+\frac{1}{y}

Step-by-step solution

Given: tan⁡A−tan⁡B=x and cot⁡B−cot⁡A=y⇒tan⁡A−tan⁡Btan⁡A⋅tan⁡B=y⇒tan⁡A⋅tan⁡B=xy for cot⁡(A−B),cot⁡(A−B)=1+tan⁡Atan⁡Btan⁡A−tan⁡B=1+xyx=(1x+1y)\begin{aligned} & \tan A-\tan B=x \\ & \text{ and } \cot B-\cot A=y \\ & \Rightarrow \frac{\tan A-\tan B}{\tan A \cdot \tan B}=y \\ & \Rightarrow \tan A \cdot \tan B=\frac{x}{y} \\ & \text{ for } \cot (A-B), \\ & \cot (A-B)=\frac{1+\tan A \tan B}{\tan A-\tan B} \\ & =\frac{1+\frac{x}{y}}{x}=\left(\frac{1}{x}+\frac{1}{y}\right) \end{aligned}
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