Q51 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
$\sin^{-1}\left(\dfrac{12}{13}\right) + \cos^{-1}\left(\dfrac{4}{5}\right) + \tan^{-1}\left(\dfrac{63}{16}\right) =$
- A$\dfrac{\pi}{2}$
- B$\dfrac{3\pi}{2}$
- Correct answerC$\pi$
- D$2\pi$
Solution
Let $A = \sin^{-1}\left(\dfrac{12}{13}\right) \Rightarrow \tan A = \dfrac{12}{5}$
Let $B = \cos^{-1}\left(\dfrac{4}{5}\right) \Rightarrow \tan B = \dfrac{3}{4}$
Let $C = \tan^{-1}\left(\dfrac{63}{16}\right)$
$\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$
$\tan(A + B) = \dfrac{\dfrac{12}{5} + \dfrac{3}{4}}{1 - \dfrac{12}{5} \times \dfrac{3}{4}} = \dfrac{\dfrac{63}{20}}{1 - \dfrac{36}{20}} = -\dfrac{63}{16}$
Since $A, B \in \left(0, \dfrac{\pi}{2}\right)$ and $\tan A \tan B = \dfrac{36}{20} > 1$, we have:
$A + B = \pi + \tan^{-1}\left(-\dfrac{63}{16}\right) = \pi - \tan^{-1}\left(\dfrac{63}{16}\right)$
$A + B = \pi - C$
$A + B + C = \pi$
Answer: $\pi$
Let $B = \cos^{-1}\left(\dfrac{4}{5}\right) \Rightarrow \tan B = \dfrac{3}{4}$
Let $C = \tan^{-1}\left(\dfrac{63}{16}\right)$
$\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$
$\tan(A + B) = \dfrac{\dfrac{12}{5} + \dfrac{3}{4}}{1 - \dfrac{12}{5} \times \dfrac{3}{4}} = \dfrac{\dfrac{63}{20}}{1 - \dfrac{36}{20}} = -\dfrac{63}{16}$
Since $A, B \in \left(0, \dfrac{\pi}{2}\right)$ and $\tan A \tan B = \dfrac{36}{20} > 1$, we have:
$A + B = \pi + \tan^{-1}\left(-\dfrac{63}{16}\right) = \pi - \tan^{-1}\left(\dfrac{63}{16}\right)$
$A + B = \pi - C$
$A + B + C = \pi$
Answer: $\pi$
Hence, the correct answer is option: 'C'.