SSC CHSL 12 Oct 2020 Shift 1 Paper

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Question : 54 of 100

Marks: +1, -0

The least value of

8 {\cosec}^{2} \θ+25 \sin ^{2} \θ

is

$10 \√{2}$
$40 \√{2}$
$20 \√{2}$
$30 \√{2}$

Solution:

\Z=8 {\cosec}^{2} \θ+25 \sin ^{2} \θ

{\dZ}/{d \θ}=16 {\cosec} \θ(-{\cosec} \θ\cot \θ) +50 \sin \θ \cos \θ

=-16 {\cos \θ}/{\sin ^{3} \θ}+50 \sin \θ\cos \θ

Now, put

{\dZ}/{d \θ}=0

, we get

-16 {\cos \θ}/{\sin ^{3} \θ}+50 \sin \θ\cos \θ=0

50 \sin \θ\cos \θ=16 {\cos \θ}/{\sin ^{3} \θ}

\sin ^{4} \θ={8}/{25}

\sin ^{2} \θ={2 \√{2}}/{5}

{\cosec}^{2} \θ={5}/{2 \√{2}}

∴

\Z=8 \×{5}/{2 \√{2}}+25 \×{2 \√{2}}/{5}

\Z=10 \√{2}+10 \√{2}

\Z=20 \√{2}

(minimum value)

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