RRB Junior Engineer Stage 1 Model Papers 4

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Question : 18 of 99

Marks: +1, -0

If

x \sin^3 θ+y \cos^3 θ=\sinθ \cos θ

and

x \sinθ=y \cos θ,\sinθ≠0,\cosθ≠0,

then

x^2+y^2

is

$1/√2$
$1/2$
1
$√2$

Solution:

x \sin^3 θ+y \cos^3 θ=\sinθ .\cos θ

⇒(x \sinθ).\sin^2θ+(y \cosθ)

cos^2θ=\sinθ.\cos θ

⇒x\sinθ.\sin^2θ+\x \sinθ.\cos^2θ

⇒\sinθ.\cosθ

⇒x \sinθ(\sin^2θ+\cos^2θ)

=\sinθ.\cosθ

⇒x=\cosθ

∴x \sinθ=y \cosθ

⇒\cos θ.\sinθ=y \cos θ

⇒y=\sinθ

∴x^2+y^2=\cos^2 θ+\sin^2 θ=1

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