BPSC 46th Exam Prelims 2004 Solved Paper

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Question : 137 of 150

Marks: +1, -0

If x = a cosθ + b sinθ and y = n sinθ – b cosθ, then

x^2 + y^2

is equal to –

$a^2 – b^2$
$b^2 – a^2$
$a^2 + b^2$
$a^2 + 2ab^2$

Solution:

Suppose that

θ=45^{∘}

From the question,

x=a \cos θ+b \sin θ

=a \cos 45^{∘}+b \sin 45^{∘}

=a × \;{1}/{√{2}}+\;{b}/{√{2}}

x^{2}=\;{a^{2}}/{2}+\;{b^{2}}/{2}+2 \;{a+b}/{√{2} × √{2}}

=\;{a^{2}}/{2}+\;{b^{2}}/{2}+a b

.....(1)
again,

y=a \sin θ-b \cos θ

y=a \sin 45^{∘}-b \cos 45^{∘}

=a \;{1}/{√{2}}-b \;{1}/{√{2}}

After squaring both side, we get:

y^{2}=(\;{a}/{√{2}})^{2}+(\;{b}/{√{2}})^{2}-2 × \;{a}/{√{2}} × \;{b}/{√{2}}

y^{2}=\;{a^{2}}/{2}+\;{b^{2}}/{2}-a b

.....(ii)
Now, adding equation (i) and (ii),

x^{2}+y^{2}=\;{a^{2}}/{2}+\;{b^{2}}/{2}+a b+\;{a^{2}}/{2}+\;{b^{2}}/{2}-a b

=a^{2}+b^{2}

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