Test Index

MHTCET 22 Apr 2025 Shift 1 Paper

© examsnet.com

Question : 125 of 150

Marks: +1, -0

The value of

\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15+\ldots \ldots \ldots \ldots \ldots+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}=

$\frac{19}{2}$
$\frac{3}{2}$
$\frac{23}{2}$
$\frac{21}{2}$

Solution:

Concept:

Use

\sin(90^\circ-\theta)=\cos\theta

and

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

to pair the terms.

Explanation:

The given sum is:

\sin^2 5^\circ+\sin^2 10^\circ+\dots+\sin^2 85^\circ+\sin^2 90^\circ

Pair each angle with its complement:

5^\circ

with

85^\circ

,

10^\circ

with

80^\circ

, and so on.

There are

8

such pairs:

\sin^2 5^\circ+\sin^2 85^\circ = \sin^2 5^\circ+\cos^2 5^\circ=1

Similarly, each of the

8

pairs gives

1

.

The unpaired middle term is

\sin^2 45^\circ

, and the last term is

\sin^2 90^\circ

.

So the total sum is:

8(1)+\sin^2 45^\circ+\sin^2 90^\circ

=8+\frac{1}{2}+1=\frac{19}{2}

Answer:

\frac{19}{2}

Option A.

© examsnet.com

Go to Question:

Prev Question

Next Question