SSC CGL 4 06 2019 Shift 1 Paper

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Question : 56 of 100

Marks: +1, -0

If

4-2 \sin ^{2} \θ-5 \cos \θ=0,0° < \θ < 90°,

then the value of

\sin \θ+\tan \θ

is

$\ {3 \√{2}}/{2}$
$\ {3 \√{3}}/{2}$
$\ {3 \√{2}}$
$\ {2 \√{3}}$

Solution:

4-2 \sin ^{2} \θ-5 \cos \θ=0

⇒

4-2\ (1-\cos ^{2} \θ\ )-5 \cos \θ=0

⇒

4-2+2 \cos ^{2} \θ-5 \cos \θ=0

⇒

2 \cos ^{2} \θ-5 \cos \θ+2=0

⇒

2 \cos ^{2} \θ-\cos \θ-4 \cos \θ+2=0

⇒

\cos \θ(2 \cos \θ-1)

-2(2 \cos \θ-1)=0

⇒

(\cos \θ-2)(2 \cos \θ-1)=0

\cos \θ=\ {1}/{2}

or

2(

not possible

)

∴

\θ=60°

now,

\sin \θ+\tan \θ

=\sin 60°+\tan 60°

=\ {\√{3}}/{2}+\√{3}=\ {3 \√{3}}/{2}

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