Test Index

UPSC NDA I 2019 Mathematics Solved Paper

© examsnet.com

Question : 48 of 120

Marks: +1, -0

What is

(1 + \tan \alpha \tan \beta)^{2} + (\tan \alpha - \tan \beta)^{2} - \sec^{2} \alpha \sec^{2} \beta

equal to?

0
1
2
4

Solution:

Concept:

Use the identity

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

and algebraic expansion formulas.

Explanation:

Step 1: Expand the first square:

(1 + \tan \alpha \tan \beta)^2 = 1 + 2\tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta

.

Step 2: Expand the second square:

(\tan \alpha - \tan \beta)^2 = \tan^2 \alpha + \tan^2 \beta - 2\tan \alpha \tan \beta

.

Step 3: Add the two expansions:

1 + 2\tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta + \tan^2 \alpha + \tan^2 \beta - 2\tan \alpha \tan \beta

= 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta

.

Step 4: Factor the expression:

1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta = (1 + \tan^2 \alpha)(1 + \tan^2 \beta)

.

Step 5: Apply the identity:

(1 + \tan^2 \alpha)(1 + \tan^2 \beta) = \sec^2 \alpha \sec^2 \beta

.

Step 6: Subtract

\sec^2 \alpha \sec^2 \beta

from the sum:

\sec^2 \alpha \sec^2 \beta - \sec^2 \alpha \sec^2 \beta = 0

.

Answer:

0

.

© examsnet.com

Go to Question:

Prev Question

Next Question