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

Option 1)
$4$

Option 2)
$2\sin 2\alpha$

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

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

Answers (1)

best_answer

As we learnt in

Formulae of Inverse Trigonometric Functions -

$\cos ^{-1}x + \cos ^{-1}y = \cos ^{-1}\left \left ( xy- \sqrt{1-x^{2}}\sqrt{1-y^{2}} \right )$

- wherein

$x\geqslant 0, y\geqslant 0$

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

$\Rightarrow \cos ^{-1} \left ( \frac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} \right ) = \alpha$

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

$= \frac{1}{2} \left \{ xy +\sqrt{1-x^2} \sqrt{4-y^2} \right \} ......... \left ( 1 \right )$

Squaring, we get

$\cos ^2\alpha = \frac{1}{4} \left [ x^{2}y^{2}+(1-x^{2})(4-y^{2})+2xy \sqrt{1-x^{2}}\sqrt{4-y^{2}} \right ]$

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

$\cos ^2\alpha = 1+ \frac{1}{4} \left [ 2x^2y^2-4x^2-y^2 \right ] + \frac{xy}{2}{\sqrt{1-x^2} \sqrt{4-y^2}}$

$\Rightarrow 1- \cos ^{2}\alpha =\frac{1}{4}\left [ 4x^2+y^2-2x^2y^2 \right ] -\frac{xy}{2}\sqrt{1-x^2}\sqrt{4-y^2}$

$\Rightarrow 4 \sin^2\alpha = 4x^2+y^2-2x^2y^2-2xy\sqrt{1-x^2}\sqrt{4-y^2} ......\left ( 2 \right )$

$= 4x^2+y^2-2x^2y^2-2xy\, \, \, \, \, \left [ 2\cos \alpha -xy \right ]$

$\Rightarrow 4 \sin ^2\alpha =4x^2+y^2-4xy\cos \alpha$

Option 1)

$4$

Incorrect

Option 2)

$2\sin 2\alpha$

Incorrect

Option 3)

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

Incorrect

Option 4)

$4\sin ^{2}\alpha$

Correct

Posted by

Aadil

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