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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2023 · Mathematics · Trigonometric Ratios & Identities

Let p=(1)/(3)-( 3 x)/( x) and q=1-3 ^2 x, 0 < x < π, x ≠ (π)/(2) ....

Let p=13−tan⁡3xtan⁡xp=\frac{1}{3}-\frac{\tan 3 x}{\tan x} and q=1−3tan⁡2x,0<x<π,x≠π2q=1-3 \tan ^2 x, 0 < x < \pi, x \neq \frac{\pi}{2}. What is pq equal to

Options

  1. A1
  2. B2
  3. C83\frac{8}{3}
  4. D−83-\frac{8}{3}

Correct answer

D. −83-\frac{8}{3}

Step-by-step solution

 Since, P=13−tan⁡3xtan⁡x=13−3tan⁡x−tan⁡3x(1−3tan⁡2x)tan⁡x=1−3tan⁡2x−9+3tan⁡2x3(1−3tan⁡2x)=−83(1−3tan⁡2x)⇒pq=−8(1−3tan⁡2x)3(1−3tan⁡2x)=−83\begin{aligned} & \text{ Since, } P=\frac{1}{3}-\frac{\tan 3 \mathrm{x}}{\tan \mathrm{x}} \\ & =\frac{1}{3}-\frac{3 \tan \mathrm{x}-\tan ^3 \mathrm{x}}{\left(1-3 \tan ^2 \mathrm{x}\right) \tan \mathrm{x}} \\ & =\frac{1-3 \tan ^2 x-9+3 \tan ^2 x}{3\left(1-3 \tan ^2 x\right)}=\frac{-8}{3\left(1-3 \tan ^2 x\right)} \\ & \Rightarrow \mathrm{pq}=\frac{-8\left(1-3 \tan ^2 \mathrm{x}\right)}{3\left(1-3 \tan ^2 \mathrm{x}\right)}=\frac{-8}{3}\end{aligned}
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