Q184 / 270 · NDA 2020 · Mathematics

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

  • A
    $\frac{K+3}{3 K-1}$
  • Correct answer
    B
    $\frac{K-3}{3 K-1}$
  • C
    $\frac{3 K-3}{K-3}$
  • D
    $\frac{K+3}{3 K+1}$

Solution

$\begin{aligned} & \text { } \frac{\tan 3 A}{\tan A}=K \\ & \frac{3 \tan A-\tan ^3 A}{\tan A \cdot\left(1-3 \tan ^2 A\right)}=K \\ & \frac{\left(3-\tan ^2 A\right)}{\left(1-3 \tan ^2 A\right)}=K \end{aligned}$ $\begin{aligned} & \Rightarrow 3-\tan ^2 A=K-3 K \tan ^2 A \\ & (3 K-1) \tan ^2 A=K-3 \Rightarrow \tan ^2 A=\frac{K-3}{(3 K-1)}\end{aligned}$

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

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