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If , then 9x^{2} – 12xy cos θ + 4y^{2} is equal to

We are given with,

We need to find the value of 9x^{2} – 12xy cos θ + 4y^{2}.

Using property of inverse trigonometry,

Take Left Hand Side (LHS) of:

Replace A by and B by .

Further solving,

We shall equate LHS to RHS,

Taking cosine on both sides,

Using property of inverse trigonometry,

cos(cos^{-1} A) = A

So,

By cross-multiplying,

⇒ xy - √(4 – x^{2}) √(9 – y^{2}) = 6 cos θ

Rearranging it,

⇒ xy – 6 cos θ = √(4 – x^{2}) √(9 – y^{2})

Squaring on both sides,

⇒ [xy – 6 cos θ]^{2} = [√(4 – x^{2}) √(9 – y^{2})]^{2}

Using algebraic identity,

(a – b)^{2} = a^{2} + b^{2} – 2ab

⇒ (xy)^{2} + (6 cos θ)^{2} – 2(xy)(6 cos θ) = (4 – x^{2})(9 – y^{2})

⇒ x^{2}y^{2} + 36 cos^{2} θ – 12xy cos θ = 36 – 9x^{2} – 4y^{2} + x^{2}y^{2}

⇒ x^{2}y^{2} – x^{2}y^{2} + 9x^{2} – 12xy cos θ + 4y^{2} = 36 – 36 cos^{2} θ

⇒ 9x^{2} – 12xy cos θ + 4y^{2} = 36 (1 – cos^{2} θ)

Using trigonometric identity,

sin^{2} θ + cos^{2} θ = 1

⇒ sin^{2} θ = 1 – cos^{2} θ

Substituting the value of (1 – cos^{2} θ), we get

⇒ 9x^{2} – 12xy cos θ + 4y^{2} = 36 sin^{2} θ

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