Test Index

MHTCET 23 Apr 2025 Shift 1 Paper

© examsnet.com

Question : 129 of 150

Marks: +1, -0

The value of

\tan ^{2}\left(\sec ^{-1} 4\right)+\cot ^{2}\left(\operatorname{cosec}^{-1} 3\right)

is

15
25
23
7

Solution:

Concept:

Use the identities

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

and

\csc^2 \theta - 1 = \cot^2 \theta

.

Explanation:

Let

\alpha = \sec^{-1} 4

, so

\sec \alpha = 4

.

Let

\beta = \csc^{-1} 3

, so

\csc \beta = 3

.

The given expression becomes

\tan^2 \alpha + \cot^2 \beta

.

Now,

\tan^2 \alpha = \sec^2 \alpha - 1 = 4^2 - 1 = 15

.

Also,

\cot^2 \beta = \csc^2 \beta - 1 = 3^2 - 1 = 8

.

Therefore,

\tan^2 \alpha + \cot^2 \beta = 15 + 8 = 23

.

Answer:

23

, which is option C.

© examsnet.com

Go to Question:

Prev Question

Next Question