SSC CGL 24 Aug 2021 Shift 3 Paper

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Question : 62 of 100

Marks: +1, -0

If

\sec 31^{∘}=x

, then

\sin ^{2} 59^{∘}+\;{1}/{\sec ^{2} 31^{∘}}

-\;{1}/{\sin ^{2} 59^{∘} cosec^{2} 59^{∘}}

is equal to

$\;{x^{2}-2}/{x^{2}}$
$\;{2-x^{2}}/{x}$
$\;{2-x^{2}}/{x^{2}}$
$\;{x^{2}-2}/{x}$

Solution:

Given,

\sec 31^{∘}=x

.
Then,

\sin ^{2} 59^{∘}+\;{1}/{\sec ^{2} 31^{∘}}-\;{1}/{\sin ^{2} 59^{∘} cosec^{2} 59^{∘}}

\;\; =\sin ^{2} 59^{∘}+\cos ^{2} 31^{∘}-\;{1}/{\sin ^{2} 59^{∘} ⋅ \;{1}/{\sin ^{2} 59^{∘}}}

\;\; =( ∵ \cos θ=\;{1}/{\sec θ})

\;\; =\sin ^{2} 59^{∘}+\cos ^{2} 31^{∘}-1

\;\; =\sin ^{2}(90^{∘}-31^{∘})+\cos ^{2} 31^{∘}-1

\;\; =\cos ^{2} 31^{∘}+\cos ^{2} 31^{∘}-1

\;\; =2 \cos ^{2} 31^{∘}-1=\;{2}/{\sec ^{2} 31^{∘}}-1

=\;{2}/{x^{2}}-1=\;{2-x^{2}}/{x^{2}}

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