SSC CGL 18 Aug 2021 Shift 3 Paper

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Question : 70 of 100

Marks: +1, -0

If

\sin ^{6} θ+\cos ^{6} θ=\;{1}/{3}, 0^{∘} < θ < 90^{∘}

, then what is the value of

\sin θ \cos θ

?

$\;{√{2}}/{3}$
$\;{√{6}}/{6}$
$\;{√{2}}/{√{3}}$
$\;{2}/{9}$

Solution:

Given,

\sin ^{6}+\cos ^{6} θ=\;{1}/{3}

⇒(\sin ^{2} θ)^{3}+(\cos ^{2} θ)^{3}=\;{1}/{3}

We know that,

(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)

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

(\sin ^{2} θ+\cos ^{2} θ)=\;{1}/{3}

⇒ 1-3 \sin ^{2} θ \cos ^{2} θ=\;{1}/{3}

⇒ \;\; 1-\;{1}/{3}=3 \sin ^{2} θ \cos ^{2} θ

⇒ \;\; \;{2}/{3}=3 \sin ^{2} θ \cos ^{2} θ

⇒ \;\; \sin ^{2} θ ⋅ \cos ^{2} θ=\;{2}/{9}

∴ \sin θ \cos θ=\;{√{2}}/{√{9}}=\;{√{2}}/{3}

.

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