SSC CHSL 10 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

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Question : 68 of 100

Marks: +1, -0

For

θ: 0^{∘} < θ < 90^{∘}

3 \sec θ+4 \cos θ=4 √{3}

, find the value of

(1-\sin θ+\cos θ)

.

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

Solution:

Given,

0^{∘} < θ < 90^{∘}

and

3 \sec θ+4 \cos θ=4 √{3}

\;{3}/{\cos θ}+4 \cos θ=4 √{3}

[ ∵ \sec θ=\;{1}/{\cos θ}]

∴ 4 \cos ^{2} θ-4 √{3} \cos θ+3=0

⇒ 4 \cos ^{2} θ-2 √{3} \cos θ-2 √{3} \cos θ

⇒ 2 \cos θ(2 \cos θ-√{3}) \;\; +3=0

\;\;\; -√{3}(2 \cos θ-√{3})=0

⇒(2 \cos θ-√{3})=0

∴ \cos θ=\;{√{3}}/{2}=\cos 30^{∘} \;\; [ ∵ 0^{∘}< θ< 90^{∘}]

∴

The value of

(1-\sin θ+\cos θ) \;\; =1-\sin 30^{∘}+\cos 30^{∘}

\;\; =1-\;{1}/{2}+\;{√{3}}/{2}

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

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