SSC CGL 20 Aug 2021 Shift 2 Paper

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Question : 59 of 100

Marks: +1, -0

If

2 \cos ^{2} θ-5 \cos θ+2=0

,

0^{∘}< θ< 90^{∘}

, then the value of

(\sec θ+\tan θ)

is

$2+√{3}$
$1-√{3}$
$1+√{3}$
$2-√{3}$

Solution:

Given,

2 \cos ^{2} θ-5 \cos θ+2=0

⇒ 2 \cos ^{2} θ-4 \cos θ-\cos θ+2=0

⇒ 2 \cos θ(\cos θ-2)-1(\cos θ-2)=0

⇒ \;\; (2 \cos θ-1)(\cos θ-2)=0

⇒ \cos θ=\;{1}/{2}, \cos θ=2 \;\;

(Not taken)

⇒ \cos θ=\cos 60^{∘}

⇒ \;\; θ=60^{∘} \;\; [ ∵ 0< θ< 90^{∘}]

∴ \sec θ+\tan θ

\;\; =\sec 60^{∘}+\tan 60^{∘}

\;\; =2+√{3}

[ ∵ \sec 60^{∘}=2

and

\tan 60^{∘}=√{3}]

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