SSC CGL 7 06 2019 Shift 2 Paper

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Question : 59 of 100

Marks: +1, -0

If

\tan ^{2} \θ-3 \sec \θ+3=0,

0° < \θ < 90°,

then the valueof

\sin \θ+\cot \θ

is:

${5√3}/6$
$2√3$
${5√3}/3$
$3√3$

Solution:

\tan ^{2} \θ-3 \sec \θ+3=0

\sec ^{2} \θ-1-3 \sec \θ+3=0

\sec ^{2} \θ-3 \sec \θ+2=0

\sec \θ=2,1

\⇒\θ=60°

as

0° < \θ < 90°

Now,

\sin \θ+\cot \θ

\⇒\ {\√{3}}/{2}+\ {1}/{\√{3}}=\ {5 \√{3}}/{6}

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