If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to

JEE Main · 2005 · 08 May 2005 · Offline · Inverse Trigonometric Functions

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

, then

4 x^{2} - 4 x y cos α + y^{2}

is equal to

Answer: (C)

Let

cos^{- 1} x = A

and

cos^{- 1} \frac{y}{2} = B

, so that

A - B = α

x = cos A

and

\frac{y}{2} = cos B

. Then

cos α = cos (A - B) = cos A cos B + sin A sin B = \frac{x y}{2} + sin A sin B

, so

sin A sin B = cos α - \frac{x y}{2}

. Squaring and using

sin^{2} A = 1 - x^{2}

and

sin^{2} B = 1 - \frac{y ^{2}}{4}

gives

(1 - x^{2}) (1 - \frac{y ^{2}}{4}) = (cos α - \frac{x y}{2})^{2}

. Expanding and multiplying through by

4

leads to

4 x^{2} - 4 x y cos α + y^{2} = 4 - 4 cos^{2} α = 4 sin^{2} α

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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