Test Index

UPSC CDS 1 2021 Math Paper

© examsnet.com

Question : 46 of 100

Marks: +1, -0

If

\sec\theta + \cos\theta = \frac{5}{2}

0 \le \theta \le 90^{\circ}

, then what is the value of

\sin^2 \theta

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
1

Solution:

Concept:

Trigonometric identity and solving a quadratic equation in

\cos \theta

.

Explanation:

Given:

\sec \theta + \cos \theta = \frac{5}{2}

, where

0 \le \theta \le 90^{\circ}

.

Rewrite

\sec \theta

as

\frac{1}{\cos \theta}

.

So

\frac{1}{\cos \theta} + \cos \theta = \frac{5}{2}

.

Multiply both sides by

2\cos \theta

:

2 + 2\cos^2 \theta = 5\cos \theta

.

Bring all terms to one side:

2\cos^2 \theta - 5\cos \theta + 2 = 0

.

Factor the quadratic:

(2\cos \theta - 1)(\cos \theta - 2) = 0

.

Thus

\cos \theta = \frac{1}{2}

or

\cos \theta = 2

.

Since

\cos \theta \le 1

for real angles,

\cos \theta = \frac{1}{2}

is the valid solution.

Then

\cos^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

.

Use the identity

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

.

Therefore

\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{4} = \frac{3}{4}

.

Answer:

\boxed{\frac{3}{4}}

(Option C).

© examsnet.com

Go to Question:

Prev Question

Next Question