SSC CGL Tier II 1 Dec 2016 Paper 1

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Question : 87 of 100

Marks: +1, -0

If

x \sin ^{3} \θ+y \cos ^{3} \θ=\sin \θ \cos \θ

and

x \sin \θ-y \cos \θ=0,

then the value of

(x^{2}+y^{2} )

equals

1
$1 /2$
$3/ 2$
2

Solution:

(1)

\∵ x \sin \θ-y \cos \θ=0

⇒ x \sin \θ=y \cos \θ..... (i)

∴x \sin ^{3} \θ+y \cos ^{3} \θ= \sin\θ . \cos\θ

⇒ y \cos \θ \sin ^{2} \θ+y \cos ^{3} \θ

=\sin \θ . \cos \θ

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

=\sin \θ .\cos \θ

⇒ y cosθ = sinθ . cosθ
⇒ y = sinθ
From equation (i),
x sinθ = sinθ . cosθ
⇒ x = cosθ
∴

x^{2}+y^{2}=\cos ^{2} \θ+\sin ^{2} \θ=1

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