SSC CGL 23 Aug 2021 Shift 2 Paper

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Question : 54 of 100

Marks: +1, -0

If

\;{\sin ^{2} θ}/{\tan ^{2} θ-\sin ^{2} θ}=5, θ

is an acute angle, then the value of

\;{24 \sin ^{2} θ-15 \sec ^{2} θ}/{6 cosec^{2} θ-7 \cot ^{2} θ}

is

$-2$
2
$-14$
14

Solution:

Given,

\;{\sin ^{2} θ}/{\tan ^{2} θ-\sin ^{2} θ}=5

⇒ \;\; \sin ^{2} θ=5 \tan ^{2} θ-5 \sin ^{2} θ

⇒ \;\; 6 \sin ^{2} θ=5 \tan ^{2} θ

⇒ \;\; 6 \sin ^{2} θ=5 × \;{\sin ^{2} θ}/{\cos ^{2} θ}

⇒ \;\; \cos ^{2} θ=\;{5}/{6}

We know that,

\sin ^{2} θ+\cos ^{2} θ=1

⇒ \;\; \sin ^{2} θ=1-\cos ^{2} θ=1-\;{5}/{6}

⇒ \;\; \sin ^{2} θ=\;{1}/{6}

Now,

\sec ^{2} θ=\;{1}/{\cos ^{2} θ}=\;{1}/{{5}/{6}}

⇒ \sec ^{2} θ=\;{6}/{5}

and

cosec^{2} θ=\;{1}/{\sin ^{2} θ}=\;{1}/{{1}/{6}}

⇒ \;\; cosec^{2} θ=6

⇒ \;\; \cot ^{2} θ=\;{\cos ^{2} θ}/{\sin ^{2} θ}=\;{\;{5}/{6}}/{{1}/{6}}

⇒ \;\; \cot ^{2} θ=5

∴ \;{24 \sin ^{2} θ-15 \sec ^{2} θ}/{6 cosec^{2} θ-7 \cot ^{2} θ}

⇒ \;{24 ×({1}/{6})-15 ×({6}/{5})}/{6 ×(6)-7 × 5}

\;\; =\;{4-18}/{36-35}=-14

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