SSC CGL 24 Aug 2021 Shift 2 Paper

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Question : 74 of 100

Marks: +1, -0

Find the value of

\sin ^{2} 60^{∘}+\cos ^{2} 30^{∘}

-\sin ^{2} 45^{∘}-3 \sin ^{2} 90^{∘}

$\;{1}/{3}$
$-1 \;{3}/{4}$
$-2 \;{1}/{2}$
$-2$

Solution:

\sin ^{2} 60^{∘}+\cos ^{2} 30^{∘}

-\sin ^{2} 45^{∘}-3 \sin ^{2} 90^{∘} .

=(\;{√{3}}/{2})^{2}+(\;{√{3}}/{2})^{2}-(\;{1}/{√{2}})^{2}-3 × 1

=\;{3}/{4}+\;{3}/{4}-\;{1}/{2}-3

[ ∵ \sin 60^{∘}=\;{√{3}}/{2}

and

\cos 60^{∘}=\;{1}/{2}

\cos 30^{∘}=\;{√{3}}/{2}

and

\sin 45^{∘}=\;{1}/{√{2}}

\sin 90^{∘}=1]

\;\; =\;{3+3-2-12}/{4}

\;\; =\;{-8}/{4}=-2

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