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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Read the following information and answer: Let ( 3 A)/(...

Paragraph: Read the following information and answer: Let tan⁡3Atan⁡A=K\frac{\tan 3 A}{\tan A}=K, where tan⁡A≠0\tan A \neq 0 and K≠13K \neq \frac{1}{3}.
Question: What is tan⁡2A\tan ^2 A equal to?

Options

  1. AK+33K−1\frac{K+3}{3 K-1}
  2. BK−33K−1\frac{K-3}{3 K-1}
  3. C3K−3K−3\frac{3 K-3}{K-3}
  4. DK+33K+1\frac{K+3}{3 K+1}

Correct answer

B. K−33K−1\frac{K-3}{3 K-1}

Step-by-step solution

tan⁡3Atan⁡A=K3tan⁡A−tan⁡3Atan⁡A⋅(1−3tan⁡2A)=K(3−tan⁡2A)(1−3tan⁡2A)=K\begin{aligned} & \frac{\tan 3 A}{\tan A}=K \\ & \frac{3 \tan A-\tan ^3 A}{\tan A \cdot\left(1-3 \tan ^2 A\right)}=K \\ & \frac{\left(3-\tan ^2 A\right)}{\left(1-3 \tan ^2 A\right)}=K \end{aligned}⇒3−tan⁡2A=K−3Ktan⁡2A(3K−1)tan⁡2A=K−3⇒tan⁡2A=K−3(3K−1)\begin{aligned} & \Rightarrow 3-\tan ^2 A=K-3 K \tan ^2 A \\ & (3 K-1) \tan ^2 A=K-3 \Rightarrow \tan ^2 A=\frac{K-3}{(3 K-1)}\end{aligned}$
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