Q225 / 270 · NDA 2023 (Phase 2) · Mathematics

Let $p=\frac{1}{3}-\frac{\tan 3 x}{\tan x}$ and $q=1-3 \tan ^2 x, 0 \lt x \lt \pi, x \neq \frac{\pi}{2}$. What is pq equal to

  • A
    1
  • B
    2
  • C
    $\frac{8}{3}$
  • Correct answer
    D
    $-\frac{8}{3}$

Solution

$\begin{aligned} & \text { Since, } P=\frac{1}{3}-\frac{\tan 3 \mathrm{x}}{\tan \mathrm{x}} \\ & =\frac{1}{3}-\frac{3 \tan \mathrm{x}-\tan ^3 \mathrm{x}}{\left(1-3 \tan ^2 \mathrm{x}\right) \tan \mathrm{x}} \\ & =\frac{1-3 \tan ^2 x-9+3 \tan ^2 x}{3\left(1-3 \tan ^2 x\right)}=\frac{-8}{3\left(1-3 \tan ^2 x\right)} \\ & \Rightarrow \mathrm{pq}=\frac{-8\left(1-3 \tan ^2 \mathrm{x}\right)}{3\left(1-3 \tan ^2 \mathrm{x}\right)}=\frac{-8}{3}\end{aligned}$

Hence, the correct answer is option: 'D'.

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