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

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

Paragraph: It is given that (θ-α)=a, (θ-β)=b Question: What is...

Paragraph:It is given that cos⁡(θ−α)=a,cos⁡(θ−β)=b\cos (\theta-\alpha)=a, \cos (\theta-\beta)=b
Question:What is cos⁡(α−β)\cos (\alpha-\beta) equal to?

Options

  1. Aab+1−a21−b2a b+\sqrt{1-a^2} \sqrt{1-b^2}
  2. Bab−1−a21−b2a b-\sqrt{1-a^2} \sqrt{1-b^2}
  3. Ca1−b2−b1−a2a \sqrt{1-b^2}-b \sqrt{1-a^2}
  4. Da1−b2+b1−a2a \sqrt{1-b^2}+b \sqrt{1-a^2}

Correct answer

A. ab+1−a21−b2a b+\sqrt{1-a^2} \sqrt{1-b^2}

Step-by-step solution

Given, cos⁡(θ−α)=a⇒sin⁡(θ−α)=1−a2\cos (\theta-\alpha)=\mathrm{a} \Rightarrow \sin (\theta-\alpha)=\sqrt{1-a^2}cos⁡(θ−β)=b⇒sin⁡(θ−β)=1−b2∴cos⁡(α−β)=cos⁡(θ−β−(θ−α))=cos⁡(θ−β)cos⁡(θ−α)+sin⁡(θ−β)sin⁡(θ−α)=(b)(a)+1−b21−a2=ab+1−a21−b2\begin{aligned} & \cos (\theta-\beta)=\mathrm{b} \Rightarrow \sin (\theta-\beta)=\sqrt{1-b^2} \\ & \therefore \cos (\alpha-\beta)=\cos (\theta-\beta-(\theta-\alpha)) \\ & =\cos (\theta-\beta) \cos (\theta-\alpha)+\sin (\theta-\beta) \sin (\theta-\alpha) \\ & \quad=(b)(a)+\sqrt{1-b^2} \sqrt{1-a^2} \\ & \quad=a b+\sqrt{1-a^2} \sqrt{1-b^2}\end{aligned}$$
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