SSC CGL 23 Aug 2021 Shift 1 Paper

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Question : 58 of 100

Marks: +1, -0

If

\sin ^{2} θ-\cos ^{2} θ-3 \sin θ+2

,

0^{∘}< θ< 90^{∘}

, then what is the value of

\;{1}/{√{\sec θ-\tan θ}}

is

$\;^{2}√{3}$
$\;^{2}√{2}$
$\;^{4}√{3}$
$\;^{4}√{2}$

Solution:

\sin ^{2} θ-\cos ^{2} θ-3 \sin θ+2=0

\sin ^{2} θ-(1-\sin ^{2} θ)-3 \sin θ+2=0

[ ∵ \sin ^{2} θ+\cos ^{2} θ=1]

⇒ \sin ^{2} θ-1+\sin ^{2} θ-3 \sin θ+2=0

⇒ \;\; 2 \sin ^{2} θ-3 \sin θ+1=0

⇒ 2 \sin ^{2} θ-2 \sin θ-\sin θ+1=0

⇒ 2 \sin θ(\sin θ-1)-1(\sin θ-1)=0

⇒ \;\; (2 \sin θ-1)(\sin θ-1)=0

⇒ \;\; \sin θ=\;{1}/{2}, \sin θ=1

and

\sin θ=\sin 30^{∘}

\sin θ=\sin 90^{∘}

θ=30^{∘}, θ=90^{∘}

θ=90^{∘}

will not take because value of

θ

varies between

0^{∘}

and

90^{∘}

.
Then,

\;\;{1}/{√{\sec θ-\tan θ}}

= \;\; \;{1}/{√{(\sec 30^{∘}-\tan 30^{∘})}}

= \;\; \;{1}/{(\;{2}/{√{3}}-\;{1}/{√{3}})^{1 \/ 2}}

\;\; =\;{1}/{[\;{1}/{√{3}}]^{1 \/ 2}}=(3)^{1 \/ 4} \;\text " or "\; \;^{4}√{3}

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