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. We will first of all assume the variables for sin A and sin B and then by calculating the value of variables of sin A and sin B. We will calculate \[{{\sin }^{2}}A\] by using \[{{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right)\] and similarly \[{{\sin }^{2}}B\] by using \[{{\sin }^{2}}B=\left( \sin B \right)\left( \sin B \right).\] And finally subtract them to get the result.

We are given to find the value of the expression

\[{{\sin }^{2}}A-{{\sin }^{2}}B......\left( i \right)\]

Let a = sin A and b = sin B. We are given \[A={{90}^{\circ }}\] and \[B={{0}^{\circ }}.\] We know that the value of \[\sin {{90}^{\circ }}=1\] and the value of \[\sin {{0}^{\circ }}=0.\]

\[\Rightarrow \sin A=\sin {{90}^{\circ }}=1\]

\[\Rightarrow \sin B=\sin {{0}^{\circ }}=0\]

Then the value of \[{{\sin }^{2}}A\] can be obtained by using \[{{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right).\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=\left( \sin {{90}^{\circ }} \right)\left( \sin {{90}^{\circ }} \right)\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\times 1\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\]

\[\Rightarrow {{\sin }^{2}}A=1\]

And

\[{{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=\left( \sin \left( {{0}^{\circ }} \right) \right)\sin {{0}^{\circ }}\]

\[\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\times 0\]

\[\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\]

Then the value of \[{{\sin }^{2}}A-{{\sin }^{2}}B\] will be

\[{{\sin }^{2}}A-{{\sin }^{2}}B={{\sin }^{2}}{{90}^{\circ }}-{{\sin }^{2}}{{0}^{\circ }}\]

\[\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1-0\]

\[\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1\]

**Complete step-by-step solution**We are given to find the value of the expression

\[{{\sin }^{2}}A-{{\sin }^{2}}B......\left( i \right)\]

Let a = sin A and b = sin B. We are given \[A={{90}^{\circ }}\] and \[B={{0}^{\circ }}.\] We know that the value of \[\sin {{90}^{\circ }}=1\] and the value of \[\sin {{0}^{\circ }}=0.\]

\[\Rightarrow \sin A=\sin {{90}^{\circ }}=1\]

\[\Rightarrow \sin B=\sin {{0}^{\circ }}=0\]

Then the value of \[{{\sin }^{2}}A\] can be obtained by using \[{{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right).\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=\left( \sin {{90}^{\circ }} \right)\left( \sin {{90}^{\circ }} \right)\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\times 1\]

\[\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\]

\[\Rightarrow {{\sin }^{2}}A=1\]

And

\[{{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=\left( \sin \left( {{0}^{\circ }} \right) \right)\sin {{0}^{\circ }}\]

\[\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\times 0\]

\[\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\]

Then the value of \[{{\sin }^{2}}A-{{\sin }^{2}}B\] will be

\[{{\sin }^{2}}A-{{\sin }^{2}}B={{\sin }^{2}}{{90}^{\circ }}-{{\sin }^{2}}{{0}^{\circ }}\]

\[\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1-0\]

\[\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1\]

**Therefore, the correct option is (a).**

**Note:**The possibility of mistake can be when the angles are \[A={{45}^{\circ }}\] or \[B={{45}^{\circ }}\] then \[\sin {{45}^{\circ }}\] would be \[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\] and \[{{\sin }^{2}}{{45}^{\circ }}=\left( \sin {{45}^{\circ }} \right)\left( \sin {{45}^{\circ }} \right)=\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{1}{2}.\] So, there can be a difference between \[{{\sin }^{2}}{{45}^{\circ }}\] and \[\sin {{45}^{\circ }}.\] Here it was \[A={{90}^{\circ }}\] and \[\sin {{90}^{\circ }}=1={{\sin }^{2}}{{90}^{\circ }},\] so that is the same here.

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