SSC CGL 17 Aug 2021 Shift 3 Paper

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Question : 73 of 100

Marks: +1, -0

The value of

\;{√{2} \tan (60^{∘}-θ) \tan (30^{∘}+θ)}/{\sin ^{2}(45^{∘}+θ)+\sin ^{2}(45^{∘}-θ)}

is

$√{2}$
2
1
$\;{1}/{√{2}}$

Solution:

\;{√{2} \tan (60^{∘}-θ) \tan (30^{∘}+θ)}/{\sin ^{2}(45^{∘}+θ)+\sin ^{2}(45^{∘}-θ)}

=\;{√{2} \tan [90^{∘}-(30^{∘}+θ)] \tan (30^{∘}+θ)}/{\sin ^{2}(45^{∘}+θ)+\sin ^{2}(45^{∘}-θ)}

[\table ∵ , \tan (90^{∘}-θ)=\cot θ ; \;\text " If "\; , \sin ^{2} A+\sin ^{2} B=1 ; \;\text " If "\; , A+B=90^{∘}]

Here,

(45^{∘}+θ+45^{∘}-θ)=90^{∘}

So,

\sin ^{2}(45^{∘}+θ)+\sin ^{2}(45^{∘}-θ)=1

\;\; =\;{√{2} \cot (30^{∘}+θ) × \tan (30^{∘}+θ)}/{1}

\;\; =\;{√{2} × 1}/{1}=√{2} . \;\; [ ∵ \cot θ × \tan θ=1]

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