SSC CGL 10 Jun 2019 Shift 2 Paper

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Question : 65 of 100

Marks: +1, -0

The value of

\sin ^{2} 60°-\cos ^{2} 45°+\sec 60°+\cos ^{2} 40°+

\cos ^{2} 50°

is equal to:

$13/4$
$7/2$
$11/2$
$9/14$

Solution:

\sin ^{2} 60°-\cos ^{2} 45°+\sec 60°

+\cos ^{2} 40°+\cos ^{2} 50°

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

+\ (\cos ^{2} 40°+\sin ^{2} 40°)

=\ {3}/{4}-\ {1}/{2}+2+1=\ {13}/{4}

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