SSC CHSL 19 Mar 2020 Shift 2 Paper

© examsiri.com

Question : 52 of 100

Marks: +1, -0

If

2 \cot \θ=3

, then

{\√{13} \cos \θ-3 \tan \θ}/{3 \tan \θ+\√{13} \sin \θ}

is

${3}/{4}$
${1}/{4}$
${2}/{3}$
${1}/{5}$

Solution:

Given,

2 \cot \θ=3

\⇒\cot \θ={3}/{2}

∴

\tan \θ={1}/{\cot \θ}={2}/{3}

{\cosec} \θ=\√{1+\cot ^{2} \θ}=\√{1+({3}/{2})^{2}}

=\√{1+{9}/{4}}=\√{{13}/{4}}={\√{13}}/{2}

\sin \θ={1}/{{\cosec} \θ}={2}/{\√{13}}

∴

{\√{13} \cos \θ-3 \tan \θ}/{3 \tan \θ+\√{13} \sin \θ}

={(\√{13} \×{3}/{\√{13}})-(3 \×{2}/{3})}/{(3 \×{2}/{3})+(\√{13} \× {2}/{\√{13}})}={3-2}/{2+2}={1}/{4}

© examsiri.com

Go to Question:

Prev Question

Next Question