Test Index

UPSC CDS I 2025 Mathematics Paper

Question : 32 of 100

Marks: +1, -0

What is

(1 + cot α−cosec α)(1 + tan α + sec α)

equal to?

Solution:

Concept:

This problem uses basic trigonometric identities:

cot \alpha = \frac{cos \alpha}{sin \alpha}

cosec \alpha = \frac{1}{sin \alpha}

tan \alpha = \frac{sin \alpha}{cos \alpha}

sec \alpha = \frac{1}{cos \alpha}

, and the identity

sin^2 \alpha + cos^2 \alpha = 1

Explanation:

Rewrite each term in parentheses using the identities:

1 + cot \alpha - cosec \alpha = 1 + \frac{cos \alpha}{sin \alpha} - \frac{1}{sin \alpha} = \frac{sin \alpha + cos \alpha - 1}{sin \alpha}

1 + tan \alpha + sec \alpha = 1 + \frac{sin \alpha}{cos \alpha} + \frac{1}{cos \alpha} = \frac{cos \alpha + sin \alpha + 1}{cos \alpha}

Multiply the two fractions:

\frac{(sin \alpha + cos \alpha - 1)(sin \alpha + cos \alpha + 1)}{sin \alpha \, cos \alpha}

Notice the numerator is of the form

(a-1)(a+1) = a^2-1

where

a = sin \alpha + cos \alpha

. So numerator becomes

(sin \alpha + cos \alpha)^2 - 1

Expand

(sin \alpha + cos \alpha)^2 = sin^2 \alpha + cos^2 \alpha + 2 sin \alpha \, cos \alpha = 1 + 2 sin \alpha \, cos \alpha

Thus numerator =

(1 + 2 sin \alpha \, cos \alpha) - 1 = 2 sin \alpha \, cos \alpha

The entire expression simplifies to

\frac{2 sin \alpha \, cos \alpha}{sin \alpha \, cos \alpha} = 2

Answer:

C. 2

Go to Question:

Prev Question

Next Question