SSC CHSL 16 Apr 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 54 of 100

Marks: +1, -0

If

(\cos ^{2} θ-1)(2 \sec ^{2} θ)+\sec ^{2} θ

+2 \tan ^{2} θ=20^{∘} < θ < 90^{∘}

, then the value of

\;{(\sec θ+\sin θ)}/{(cosec θ-\cos θ)}

will be

$-1$
$-3$
3
2

Solution:

Given,

(\cos ^{2} θ-1)(2 \sec ^{2} θ)

+\sec ^{2} θ+2 \tan ^{2} θ=2

(\cos ^{2} θ-1) × \;{2}/{\cos ^{2} θ}+\tan ^{2} θ +1

+2 \tan ^{2} θ=2

\;2-2 \sec ^{2} θ+1+3 \tan ^{2} θ=2

\;\; -2 \tan ^{2} θ-2+1+3 \tan ^{2} θ=0

∴ \;\; \tan ^{2} θ=1 ⇒ \tan θ=1=\tan 45^{∘}

∴ \;\; θ \;\; =45^{∘} \;\; [ ∵ 0^{∘} < θ < 90^{∘}]

\;\text " Now, the value of "\;

\;{\sec 45^{∘}+\sin 45^{∘}}/{cosec 45^{∘}-\cos 45^{∘}}

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

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