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MHTCET 18 Apr 2026 Shift 2 Paper

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Question : 118 of 150

Marks: +1, -0

\sec ^{2}\left(\tan ^{-1} 3\right)-\tan ^{2}\left(\sec ^{-1} 3\right)=

0
1
2
3

Solution:

Concept:

Use inverse trigonometric definitions and Pythagorean identities to simplify the terms.

Explanation:

Let

\theta = \tan^{-1} 3

.

Then

\tan \theta = 3

.

Using

\sec^2 \theta = 1 + \tan^2 \theta

, we get:

\sec^2(\tan^{-1} 3) = 1 + 3^2 = 10

.

Let

\phi = \sec^{-1} 3

.

Then

\sec \phi = 3

.

Using

\tan^2 \phi = \sec^2 \phi - 1

, we get:

\tan^2(\sec^{-1} 3) = 3^2 - 1 = 8

.

Now evaluate the expression:

\sec^2(\tan^{-1} 3) - \tan^2(\sec^{-1} 3) = 10 - 8 = 2

.

Answer:

2

i.e. Option C.

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