Q66 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
$\sec^2(\tan^{-1}3) - \tan^2(\sec^{-1}3) = $
- A$0$
- B$1$
- Correct answerC$2$
- D$3$
Solution
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$, we get:
$\sec^2(\tan^{-1} 3) = 1 + \tan^2(\tan^{-1} 3) = 1 + (3)^2 = 10$
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, we get:
$\tan^2(\sec^{-1} 3) = \sec^2(\sec^{-1} 3) - 1 = (3)^2 - 1 = 8$
Substituting these values into the given expression:
$\sec^2(\tan^{-1} 3) - \tan^2(\sec^{-1} 3) = 10 - 8 = 2$
$\sec^2(\tan^{-1} 3) = 1 + \tan^2(\tan^{-1} 3) = 1 + (3)^2 = 10$
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, we get:
$\tan^2(\sec^{-1} 3) = \sec^2(\sec^{-1} 3) - 1 = (3)^2 - 1 = 8$
Substituting these values into the given expression:
$\sec^2(\tan^{-1} 3) - \tan^2(\sec^{-1} 3) = 10 - 8 = 2$
Hence, the correct answer is option: 'C'.