If we have an expression as ${{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha $, where $-1\le x\le 1,-2\le y\le 2,x\le \dfrac{y}{2}$ the $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to
[a] $4{{\sin }^{2}}\alpha +2{{x}^{2}}{{y}^{2}}$
[b] $4{{\sin }^{2}}\alpha -2{{x}^{2}}{{y}^{2}}$
[c] $4{{\sin }^{2}}\alpha $
[d] None of these

Hint: Take cos on both sides of the equation ${{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha $ and use the fact that $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$. Simplify and hence find the value of the expression $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$.

Complete step-by-step solution:
We have ${{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha $
Taking cos on both sides, we get
$\cos \left( {{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2} \right)=\cos \alpha $
We know that $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
Hence, we have
$\cos \left( {{\cos }^{-1}}x \right)\cos \left( {{\cos }^{-1}}\dfrac{y}{2} \right)+\sin \left( {{\cos }^{-1}}x \right)\sin \left( {{\cos }^{-1}}\dfrac{y}{2} \right)=\cos \alpha $
We know that $\cos \left( {{\cos }^{-1}}x \right)=x\forall x\in \left[ -1,1 \right]$ and $\sin \left( {{\cos }^{-1}}x \right)=\sqrt{1-{{x}^{2}}}\forall x\in \left[ -1,1 \right]$
Hence, we have
$x\dfrac{y}{2}+\sqrt{1-{{x}^{2}}}\times \sqrt{1-\dfrac{{{y}^{2}}}{4}}=\cos \alpha $
Subtracting $\dfrac{xy}{2}$ on both sides, we get
$\sqrt{1-{{x}^{2}}}\dfrac{\sqrt{4-{{y}^{2}}}}{2}=\cos \alpha -\dfrac{xy}{2}$
Multiplying by 2 on both sides, we get
$\sqrt{1-{{x}^{2}}}\sqrt{4-{{y}^{2}}}=2\cos \alpha -xy$
Squaring both sides, we get
$\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)={{\left( 2\cos \alpha -xy \right)}^{2}}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, we have
$\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)=4{{\cos }^{2}}\alpha +{{x}^{2}}{{y}^{2}}-4xy\cos \alpha $
Expanding the term on LHS, we get
$4-{{y}^{2}}-4{{x}^{2}}+{{x}^{2}}{{y}^{2}}=4{{\cos }^{2}}\alpha -4xy\cos \alpha +{{x}^{2}}{{y}^{2}}$
Subtracting ${{x}^{2}}{{y}^{2}}$ from both sides, we get
$4-{{y}^{2}}-4{{x}^{2}}=4{{\cos }^{2}}\alpha -4xy\cos \alpha $
Adding $4{{x}^{2}}+{{y}^{2}}$ on both sides, we get
$4=4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha $
Subtracting $4{{\cos }^{2}}\alpha $ on both sides, we get
$4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4-4{{\cos }^{2}}\alpha =4\left( 1-{{\cos }^{2}}\alpha \right)$
We know that $1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha $
Hence, we have
$4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha $
Hence option [c] is correct.

Note: Alternative solution:
We know that ${{\cos }^{-1}}x-{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)+2k\pi ,k\in \mathbb{N}$, where k is suitably chosen.
Hence, we have
$\begin{align}
  & {{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}={{\cos }^{-1}}\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right)+2k\pi ,k\in \mathbb{N} \\
 & \Rightarrow \alpha ={{\cos }^{-1}}\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right)+2k\pi ,k\in \mathbb{N} \\
\end{align}$
Taking cos on both sides, we get
$\cos \alpha =\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right)$, which is the same as obtained above. Hence proceeding as above, we get option [c] is correct.