Question

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

Answer 1

. 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 $${\sin }^{2}A=(\sin A)(\sin A)$$ and similarly $${\sin }^{2}B$$ by using $${\sin }^{2}B=(\sin B)(\sin B).$$ 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 $${\sin }^{2}A=(\sin A)(\sin A).$$
$$\Rightarrow {\sin }^{2}A={\sin }^2{90}^{\circ }=(\sin {90}^{\circ })(\sin {90}^{\circ })$$
$$\Rightarrow {\sin }^{2}A={\sin }^2{90}^{\circ }=1\times 1$$
$$\Rightarrow {\sin }^{2}A={\sin }^2{90}^{\circ }=1$$
$$\Rightarrow {\sin }^{2}A=1$$
And
$${\sin }^{2}B={\sin }^2{0}^{\circ }=(\sin \left(0^{\circ }\right))\sin 0^{\circ }$$
$$\Rightarrow {\sin }^{2}B={\sin }^2{0}^{\circ }=0\times 0$$
$$\Rightarrow {\sin }^{2}B={\sin }^2{0}^{\circ }=0$$
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 $$\sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$$ and $${\sin }^2{45}^{\circ }=(\sin {45}^{\circ })(\sin {45}^{\circ })=\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\displaystyle \frac{1}{2}}.$$ 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.