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# If ${{\cos }^{-1}}x-{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha$ then $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to?\begin{align} & \left( A \right)4{{\sin }^{2}}\alpha \\ & \left( B \right)-4{{\sin }^{2}}\alpha \\ & \left( C \right)2\sin 2\alpha \\ & \left( D \right)4 \\ \end{align}

Last updated date: 09th Aug 2024
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Hint: We first apply the formula of ${{\cos }^{-1}}x+{{\cos }^{-1}}y$ on the left hand side of the given equation. We then simplify the equation, and square it. Rearranging the terms and applying some basic trigonometric formulae, we get $\left( A \right)$ as the correct option.

The given equation is
${{\cos }^{-1}}x+{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha$
We know the formula that ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)$ . Thus, applying this formula in the above equation the equation thus becomes,
$\Rightarrow {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right)=\alpha$
Taking $\text{cosine}$ on both sides on the above equation, we get,
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right) \right)=\cos \alpha$
We know the simple formula that $\cos \left( {{\cos }^{-1}}x \right)=\cos x$ . So, applying this in the above equation, the equation thus becomes,
$\Rightarrow x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)}=\cos \alpha$
Simplifying the above equation, we get,
$\Rightarrow \dfrac{xy}{2}+\sqrt{\left( 1-{{x}^{2}} \right)\left( \dfrac{4-{{y}^{2}}}{4} \right)}=\cos \alpha$
Further simplifying the above equation, the equation thus becomes,
$\Rightarrow \dfrac{xy}{2}+\dfrac{\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}}{2}=\cos \alpha$
Multiplying both sides of the above equation by $2$ , we get,
$\Rightarrow xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}=2\cos \alpha$
Taking $\cos \alpha$ to the left hand side of the above equation and $\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}$ to the right hand side of the above equation, we get,
$\Rightarrow xy-2\cos \alpha =-\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}$
Squaring both sides of the above equation, the equation thus becomes,
$\Rightarrow {{\left( xy-2\cos \alpha \right)}^{2}}={{\left( -\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)} \right)}^{2}}$
Evaluating the above equation, we get
$\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)$
Opening the brackets in the above equation, we get,
$\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}$
Subtracting ${{x}^{2}}{{y}^{2}}$ from both sides of the above equation, we get,
$\Rightarrow -4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}$
Bringing the terms $4{{x}^{2}},{{y}^{2}}$ to the left hand side of the above equation and the term $4{{\cos }^{2}}\alpha$ to the right hand side of the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4-4{{\cos }^{2}}\alpha$
Taking $4$ common in the right hand side of the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4\left( 1-{{\cos }^{2}}\alpha \right)$
We know that $1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha$ . Thus, applying this formula in the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4{{\sin }^{2}}\alpha$
Rearranging the terms of the above equation, we get,
$\Rightarrow 4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha$
This is nothing but the thing that we have to prove. Therefore, we can conclude that $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to $4{{\sin }^{2}}\alpha$ which is option $\left( A \right)$ .

Note: We must be very careful while carrying out the square as this expression deals with a little complex terms and students are prone to make mistakes here. This problem can also be solved by taking some values for $x,y,\alpha$ and find out which of the following options gives the correct answer. Let’s take $x=0,y=1$ . Then, $\alpha$ becomes ${{\cos }^{-1}}0-{{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}-\dfrac{\pi }{3}=\dfrac{\pi }{6}$ . The expression becomes $4{{\left( 0 \right)}^{2}}-4\left( 0 \right)\left( 1 \right)\cos \left( \dfrac{\pi }{6} \right)+{{\left( 1 \right)}^{2}}=1$ . Out of the following options, only $\left( A \right)$ satisfies by putting $\alpha =\dfrac{\pi }{6}$.