Q192 / 270 · NDA 2024 (Phase 1) · Mathematics

What is $\sqrt{15+\cot ^2\left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)}$ equal to?

  • A
    1
  • B
    7
  • Correct answer
    C
    8
  • D
    16

Solution

$\begin{aligned} 2 \cot ^{-1} 3 & =2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}} \\ & =\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)=\tan ^{-1} \frac{3}{4}\end{aligned}$ So, $\begin{aligned} \sqrt{15+\cot ^2\left(\frac{\pi}{4}\right.}- & \left.2 \cot ^{-1} 3\right) \\ & =\sqrt{15+\cot ^2\left(\tan ^{-1} 1-\tan ^{-1} \frac{3}{4}\right)} \\ & =\sqrt{15+\cot ^2\left(\tan ^{-1}\left(\frac{\left.1-\frac{3}{4}\right)}{1+\frac{3}{4}}\right)\right]} \\ & =\sqrt{15+\cot ^2\left(\tan ^{-1} \frac{1}{7}\right)} \\ & =\sqrt{15+7^2} \\ & =\sqrt{15+\cot ^2\left(\cot ^{-1} 7\right)} \\ & =\sqrt{15+49}=8 \end{aligned}$

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

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