Q66 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
The value of $\dfrac{\tan^{-1}(1) + \cos^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{1}{2}\right)}{\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)}$ is equal to
- A$\dfrac{3}{4}$
- B$\dfrac{-1}{3}$
- C$-4$
- Correct answerD$\dfrac{-9}{4}$
Solution
The given expression is $\dfrac{\tan^{-1}(1) + \cos^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{1}{2}\right)}{\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)}$
Evaluating the terms in the numerator:
$\tan^{-1}(1) = \dfrac{\pi}{4}$
Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \dfrac{\pi}{2}$, we have $\cos^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{2}$
Numerator $= \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}$
Evaluating the terms in the denominator:
$\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$
$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$
Denominator $= \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$
Substituting these values back into the expression:
$\dfrac{\dfrac{3\pi}{4}}{-\dfrac{\pi}{3}} = \dfrac{3\pi}{4} \times \left(-\dfrac{3}{\pi}\right) = -\dfrac{9}{4}$
Answer: $\dfrac{-9}{4}$
Evaluating the terms in the numerator:
$\tan^{-1}(1) = \dfrac{\pi}{4}$
Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \dfrac{\pi}{2}$, we have $\cos^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{2}$
Numerator $= \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}$
Evaluating the terms in the denominator:
$\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$
$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$
Denominator $= \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$
Substituting these values back into the expression:
$\dfrac{\dfrac{3\pi}{4}}{-\dfrac{\pi}{3}} = \dfrac{3\pi}{4} \times \left(-\dfrac{3}{\pi}\right) = -\dfrac{9}{4}$
Answer: $\dfrac{-9}{4}$
Hence, the correct answer is option: 'D'.