SSC CPO SI and ASI 2012 Paper

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Question : 139 of 200

Marks: +1, -0

If cosec θ – sin θ = l and sec θ – cos θ = m, then the value of l

^{2}

m

^{2}

(l

^{2 }

+ m

^{2}

+ 3) is

–1
0
1
2

Solution:

(3)Here,

l^{2} m^{2}=(\cosec \θ-\sin \θ)^{2}(\sec \θ-\cos \θ)^{2}

=\ (\ {1}/{\sin \θ}-\sin \θ)^{2}\ (\ {1}/{\cos \θ}-\cos \θ\ )^{2}

=\ {\cos ^{4} \θ}/{\sin ^{2} \θ} \×{\sin ^{4} θ}/{\cos ^{2} \θ}=\sin ^{2} \θ \cos ^{2} \θ

And,

l^{2}+m^{2}+3=\ {\cos ^{4} \θ}{\sin ^{2} \θ}+ {\sin ^{4} \θ}/{\cos ^{2} \θ}+3

=\ {\cos ^{6} \θ+\sin ^{6} \θ+3 \sin ^{2} \θ \cos ^{2} \θ}/{\sin ^{2} \θ \cos ^{2} \θ }

=\ {\ (\cos ^{2} θ+\sin ^{2} θ)^{3}-3 \sin ^{2} θ \cos ^{2} θ\ (\cos ^{2} θ+\sin ^{2} θ )+3 \sin ^{2} θ \cos ^{2} θ}/{\sin ^{2} θ \cos ^{2} θ}

=\ {1}/{\sin ^{2} \θ \cos ^{2} \θ} \sin ^{2} \θ+\cos ^{2} \θ=1

Therefore,

\ ( l^{2} m^{2} ) (l^{2}+m^{2}+3\ )=1

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