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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 42 of 120

Marks: +1, -0

What is the value of the following determinant?

\begin{vmatrix} \cos \mathrm{C} & \tan \mathrm{A} & 0 \\ \sin \mathrm{B} & 0 & -\tan \mathrm{A} \\ 0 & \sin \mathrm{B} & \cos \mathrm{C} \end{vmatrix}

$-1$
0
2tan A sin B sin C
-2tan A sin B sin C

Solution:

Concept:

The value of a

3 \times 3

determinant is found by expanding along any row or column using the formula for minors and cofactors.

Explanation:

Let

D = \begin{vmatrix} \cos C & \tan A & 0 \\ \sin B & 0 & -\tan A \\ 0 & \sin B & \cos C \end{vmatrix}

.

Expand

D

along the first row:

D = \cos C \cdot \begin{vmatrix} 0 & -\tan A \\ \sin B & \cos C \end{vmatrix} - \tan A \cdot \begin{vmatrix} \sin B & -\tan A \\ 0 & \cos C \end{vmatrix} + 0 \cdot \begin{vmatrix} \sin B & 0 \\ 0 & \sin B \end{vmatrix}

.

Compute each

2 \times 2

determinant:

First term:

\cos C \cdot (0 \cdot \cos C - (-\tan A) \cdot \sin B) = \cos C \cdot (\tan A \sin B) = \tan A \sin B \cos C

.

Second term:

-\tan A \cdot (\sin B \cdot \cos C - (-\tan A) \cdot 0) = -\tan A \cdot (\sin B \cos C) = -\tan A \sin B \cos C

.

Third term:

0

.

Add the results:

\tan A \sin B \cos C - \tan A \sin B \cos C = 0

.

Answer:

0

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