If cos -1x-3-cos -1y-2-2 then 4 x2-12 x y cos-2-9 y2-

The correct option is D 18 18 cos θcos θ/2=xy/6 √((1 x^2/9 )(1 y^2/4 ))⇒(cos θ/2 xy/6 )^2=1 x^2/9 y^2/4+x^2y^2/36 ...

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If cos -1x/3+...

Question

If $c o s^{- 1} (\frac{x}{3}) + c o s^{- 1} (\frac{y}{2}) = \frac{θ}{2} t h e n 4 x^{2} - 12 x y c o s \frac{θ}{2} + 9 y^{2} =$

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36 - 36 c o s θ

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18 - 18 c o s θ

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18 + 18 c o s θ

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{cos}^{- 1} \frac{x}{3} + {cos}^{- 1} \frac{y}{2} = \frac{θ}{2}

4 x^{2} - 12 x y cos \frac{θ}{2} + 9 y^{2} is equal to:

{cos}^{- 1} (\frac{x}{2}) + {cos}^{- 1} (\frac{y}{3}) = θ

9 x^{2} - 12 x y cos θ + 4 y^{2}

{cos}^{- 1} (\frac{x}{2}) + {cos}^{- 1} (\frac{y}{3}) = θ,

9 x^{2} - 12 x y cos θ + 4 y^{2} = m {sin}^{2} θ .

m

c o s^{- 1} x - c o s^{- 1} (\frac{y}{2}) = α

4 x^{2} - 4 x y cos α + y^{2} =

{cos}^{- 1} x - {cos}^{- 1} \frac{y}{2} = α

- 1 \leq x \leq 1,

- 2 \leq y \leq 2, x \leq \frac{y}{2},

x, y, 4 x^{2} - 4 x y cos α + y^{2}

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