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# When the value of angle A is ${{90}^{\circ }}$ and B is ${{0}^{\circ }}$ then find the value of ${{\sin }^{2}}A-{{\sin }^{2}}B.$$\left( a \right)0$$\left( b \right)\dfrac{1}{2}$$\left( c \right)1$$\left( d \right)2$

Last updated date: 19th Sep 2024
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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.

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.