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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

If f(α)= ^2 α-1 , then what is (f(α)+f(β))/(1-f(α) f(β)) equal to?

If f(α)=sec⁡2α−1f(\alpha)=\sqrt{\sec ^2 \alpha-1}, then what is f(α)+f(β)1−f(α)f(β)\frac{f(\alpha)+f(\beta)}{1-f(\alpha) f(\beta)} equal to?

Options

  1. Af(α−β)f(\alpha-\beta)
  2. Bf(α+β)f(\alpha+\beta)
  3. Cf(α)f(β)f(\alpha) f(\beta)
  4. Df(αβ)f(\alpha \beta)

Correct answer

B. f(α+β)f(\alpha+\beta)

Step-by-step solution

f(α)=sec⁡2α−1=tan⁡α∴f(α)+f(β)1−f(α)⋅f(β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=tan⁡(α+β)=f(α+β)\begin{aligned} & f(\alpha)=\sqrt{\sec ^2 \alpha-1}=\tan \alpha \\ & \therefore \frac{f(\alpha)+f(\beta)}{1-f(\alpha) \cdot f(\beta)}=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \\ & =\tan (\alpha+\beta)=f(\alpha+\beta)\end{aligned}
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