SSC CPO SI and ASI 2014 Paper

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Question : 141 of 200

Marks: +1, -0

If 0° < θ < 90° and 2 sec θ = 3 cosec

^{2}

θ, then θ is

$π/ 6$
$π/ 4$
$π/ 3$
$π/ 5$

Solution:

(3) Given, 2 sec θ = 3 cosec

^{2}

θ

{2}/{\cos \θ}= {3}/{\sin ^{2} \θ}

⇒{2}/{\cos \θ}= {3}/{1-\cos ^{2} \θ}

2 \cos ^{2} \θ+3 \cos \θ-2

=2 \cos ^{2} \θ+4 \cos \θ-\cos \θ-2=0

2 \cos \θ(\cos \θ+2)-1(\cos \θ+2)=0

and

(2 \cos \θ-1)(\cos \θ+2)=0

So,

\cos \θ= {1}/{2}=\cos 60° \θ= {\π}/{3}

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