SSC CGL 11 Jun 2019 Shift 1 Paper

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Question : 62 of 100

Marks: +1, -0

The value of

\sec ^{2} 28^°-\cot ^{2} 62°+\sin ^{2} 60°

+\ {\cosec}^{2} 30°

isequal to:

$7/2$
3
$19/4$
$23/4$

Solution:

\sec ^{2} 28°-\cot ^{2} 62°+\sin ^{2} 60°+\ {\cosec}^{2} 30°

=\ {\Sec}^{2} 28°-\tan ^{2} 28°+\ {3}/{4}+4

=1+\ {3}/{4}+4=\ {4+3+16}/{4}

=\ {23}/{4}

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