Q196 / 270 · NDA 2025 (Phase 2) · Mathematics
What is $\cot^{-1} 9 + \text{cosec}^{-1}\left(\dfrac{\sqrt{41}}{4}\right)$ equal to?
- Correct answerA$\dfrac{\pi}{4}$
- B$\dfrac{\pi}{3}$
- C$\dfrac{\pi}{2}$
- D$\pi$
Solution
Let $A = \cot^{-1} 9 \Rightarrow \cot A = 9 \Rightarrow \tan A = \dfrac{1}{9}$
Let $B = \text{cosec}^{-1}\left(\dfrac{\sqrt{41}}{4}\right) \Rightarrow \text{cosec} B = \dfrac{\sqrt{41}}{4} \Rightarrow \sin B = \dfrac{4}{\sqrt{41}}$
From this, $\tan B = \dfrac{4}{\sqrt{(\sqrt{41})^2 - 4^2}} = \dfrac{4}{\sqrt{41 - 16}} = \dfrac{4}{5}$
Using the formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan(A + B) = \dfrac{\dfrac{1}{9} + \dfrac{4}{5}}{1 - \left(\dfrac{1}{9}\right)\left(\dfrac{4}{5}\right)}$
$\tan(A + B) = \dfrac{\dfrac{5 + 36}{45}}{\dfrac{45 - 4}{45}}$
$\tan(A + B) = \dfrac{41}{41} = 1$
Since $A, B \in \left(0, \dfrac{\pi}{2}\right)$, we have $A + B = \dfrac{\pi}{4}$
Answer: $\dfrac{\pi}{4}$
Let $B = \text{cosec}^{-1}\left(\dfrac{\sqrt{41}}{4}\right) \Rightarrow \text{cosec} B = \dfrac{\sqrt{41}}{4} \Rightarrow \sin B = \dfrac{4}{\sqrt{41}}$
From this, $\tan B = \dfrac{4}{\sqrt{(\sqrt{41})^2 - 4^2}} = \dfrac{4}{\sqrt{41 - 16}} = \dfrac{4}{5}$
Using the formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan(A + B) = \dfrac{\dfrac{1}{9} + \dfrac{4}{5}}{1 - \left(\dfrac{1}{9}\right)\left(\dfrac{4}{5}\right)}$
$\tan(A + B) = \dfrac{\dfrac{5 + 36}{45}}{\dfrac{45 - 4}{45}}$
$\tan(A + B) = \dfrac{41}{41} = 1$
Since $A, B \in \left(0, \dfrac{\pi}{2}\right)$, we have $A + B = \dfrac{\pi}{4}$
Answer: $\dfrac{\pi}{4}$
Hence, the correct answer is option: 'A'.