Q22 / 120 · NDA 2024 (Phase 2) · Mathematics

What is $\cot ^2\left(\sec ^{-1} 2\right)+\tan ^2\left(\operatorname{cosec}^{-1} 3\right)$ equal to?

  • A
    $\frac{11}{12}$
  • Correct answer
    B
    $\frac{11}{24}$
  • C
    $\frac{7}{24}$
  • D
    $\frac{1}{24}$

Solution

$\begin{aligned} & \text { Suppose } \mathrm{I}=\cot ^2\left(\sec ^{-1} 2\right)+\tan ^2\left(\operatorname{cosec}^{-1} 3\right) \\ & \Rightarrow \mathrm{I}=\cot ^2\left(60^{\circ}\right)+\tan ^2+\left(\cot ^{-1} \sqrt{9-1}\right) \\ & \left\{\therefore \operatorname{cosec}^{-1} \mathrm{t}=\cot ^{-1} \sqrt{\mathrm{t}^2-1}\right\} \\ & \Rightarrow \mathrm{I}=\frac{1}{3}+\tan ^2\left(\tan ^{-1} \frac{1}{\sqrt{8}}\right)=\frac{1}{3}+\frac{1}{8}=\frac{11}{24}\end{aligned}$

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

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