If $\cos^{-1}x-\cos^{-1}\frac{y}{2}=\alpha$ , where $-1\leq x\leq 1$ ,

$-2\leq y\leq 2$ , $x\leq \frac{y}{2}$ , then for all $x,y,4x^{2}-4xy\cos \alpha +y^{2}$

is equal to :

Option 1)
$4\sin ^{2}\alpha$

Option 2)
$2\sin ^{2}\alpha$

Option 3)
$4\sin ^{2}\alpha-2x^{2}y^{2}$

Option 4)
$4\cos ^{2}\alpha+2x^{2}y^{2}$

Answers (1)

P Plabita

$\cos^{-1}x-\cos^{-1}\frac{y}{2}=\alpha$ where $-1\leq x\leq 1$ , $-2\leq y\leq 2$ , $x\leq \frac{y}{2}$

$\cos \alpha =\cos (\cos^{-1}x-\cos^{-1}\frac{y}{2})$

$=\cos (\cos^{-1}x)\cdot cos(\cos^{-1}\frac{y}{2})+sin(\cos^{-1}x)sin(\cos^{-1}\frac{y}{2})$

$=\frac{xy}{2}+\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}$

$cos\alpha -\frac{xy}{2}=\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}$

Squaring both the sides

$cos^{2}\alpha +\frac{x^{2}y^{2}}{4}-xycos\alpha =(1-x^{2})({1-\frac{y^{2}}{4}})$

$4-4x^{2}-y^{2}=4cos^{2}\alpha -4xycos\alpha$

$=4(1-cos^{2}\alpha )$

$=4sin^{2}\alpha$

So, option (1) is correct.

Option 1)

$4\sin ^{2}\alpha$

Option 2)

$2\sin ^{2}\alpha$

Option 3)

$4\sin ^{2}\alpha-2x^{2}y^{2}$

Option 4)

$4\cos ^{2}\alpha+2x^{2}y^{2}$

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If , where , , , then for all is equal to : Option 1) Option 2) Option 3) Option 4)

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