SSC CGL 24 Aug 2021 Shift 3 Paper

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Question : 59 of 100

Marks: +1, -0

Simplify

\;{(\sin θ+\sec θ)^{2}+(\cos θ+cosec θ)^{2}}/{(1+\sec θ cosec θ)^{2}},

0^{∘} < θ < 90^{∘}

.

0
1
$-1$
2

Solution:

Given,

\;{(\sin θ+\sec θ)^{2}+(\cos θ+cosec θ)^{2}}/{(1+\sec θ cosec θ)^{2}} \;\text ", "\;

[ ∵ 0^{∘} < θ < 90^{∘}]

(\sin ^{2} θ+\sec ^{2} θ+2 \sin θ \sec θ+\cos ^{2} θ

=\;{+cosec^{2} θ+2 \cos θ cosec θ}/{(1+\sec θ cosec θ)^{2}}

(\sin ^{2} θ+\cos ^{2} θ)+(\sec ^{2} θ+2 \sin θ \sec θ)

=\;{+(cosec/{ }^{2} θ+2 \cos θ cosec θ)}{(1+\sec θ cosec^{2} θ)}

=\;{1+({1}/{\cos ^{2} θ}+2 \tan θ)+({1}/{\sin ^{2} θ}+2 \cot θ)}/{(1+\sec θ × cosec θ)^{2}}

=\;{[1+{1}/{\cos ^{2} θ}+{1}/{\sin ^{2} θ}+2({\sin θ}/{\cos θ}+{\cos θ}/{\sin θ})]}/{(1+\sec θ cosec θ)^{2}}

=\;{1+{(\sin ^{2} θ+\cos ^{2} θ)}/{\cos ^{2} θ \sin ^{2} θ}+2 × {(\sin ^{2} θ+\cos ^{2} θ)}/{\sin θ \cos θ}}/{(1+\sec θ cosec θ)^{2}}

=\;{(1+{1}/{\cos ^{2} θ \sin ^{2} θ}+2 × {(1)}/{\sin θ \cos θ})}/{(1+\sec θ cosec θ)^{2}}

=\;{(1+cosec^{2} θ \sec ^{2} θ+2 cosec θ \sec θ)}/{(1+\sec θ cosec θ)^{2}}

=\;{(1+cosec θ \sec θ)^{2}}/{(1+cosec θ \sec θ)^{2}}=1

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