SSC CHSL 15 Oct 2020 Shift 1 Paper

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Question : 51 of 100

Marks: +1, -0

Quantitative Aptitude

{\cos x}/{1+\sin x}+{1+\sin x}/{\cos x}

is equal to

$2 \sin x$
$2 \cos x$
$2 {\cosec} x$
$2 \sec x$

Solution:

Given,

{\cos x}/{1+\sin x}+{1+\sin x}/{\cos x}

={\cos x}/{1+\sin x} ×{1-\sin x}/{1-\sin x}+{1+\sin x}/{\cos x}

={\cos x ×(1-\sin x)}/{(1-\sin ^{2} x)}+{1}/{\cos x}+{\sin x}/{\cos x}

[\∵ (a+b)(a-b)=a^{2}-b^{2}]

={\cos x(1-\sin x)}/{\cos ^{2} x}+\sec x+\tan x

[\∵ \sin ^{2} x+\cos ^{2} x=1]

={1}/{\cos x}-{\sin x}/{\cos x}+\sec x+\tan x

=\sec x-\tan x+\sec x+\tan x

=2 \sec x

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