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Question: When the value of angle A is \[{{90}^{\circ }}\] and B is \[{{0}^{\circ }}\] then find the value of ...

When the value of angle A is 90∘{{90}^{\circ }} and B is 0∘{{0}^{\circ }} then find the value of sin⁡2A−sin⁡2B.{{\sin }^{2}}A-{{\sin }^{2}}B.
(a)0\left( a \right)0
(b)12\left( b \right)\dfrac{1}{2}
(c)1\left( c \right)1
(d)2\left( d \right)2

Explanation

Solution

. 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⁡2A{{\sin }^{2}}A by using sin⁡2A=(sin⁡A)(sin⁡A){{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right) and similarly sin⁡2B{{\sin }^{2}}B by using sin⁡2B=(sin⁡B)(sin⁡B).{{\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⁡2A−sin⁡2B......(i){{\sin }^{2}}A-{{\sin }^{2}}B......\left( i \right)
Let a = sin A and b = sin B. We are given A=90∘A={{90}^{\circ }} and B=0∘.B={{0}^{\circ }}. We know that the value of sin⁡90∘=1\sin {{90}^{\circ }}=1 and the value of sin⁡0∘=0.\sin {{0}^{\circ }}=0.
⇒sin⁡A=sin⁡90∘=1\Rightarrow \sin A=\sin {{90}^{\circ }}=1
⇒sin⁡B=sin⁡0∘=0\Rightarrow \sin B=\sin {{0}^{\circ }}=0
Then the value of sin⁡2A{{\sin }^{2}}A can be obtained by using sin⁡2A=(sin⁡A)(sin⁡A).{{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right).
⇒sin⁡2A=sin⁡290∘=(sin⁡90∘)(sin⁡90∘)\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=\left( \sin {{90}^{\circ }} \right)\left( \sin {{90}^{\circ }} \right)
⇒sin⁡2A=sin⁡290∘=1×1\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\times 1
⇒sin⁡2A=sin⁡290∘=1\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1
⇒sin⁡2A=1\Rightarrow {{\sin }^{2}}A=1
And
sin⁡2B=sin⁡20∘=(sin⁡(0∘))sin⁡0∘{{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=\left( \sin \left( {{0}^{\circ }} \right) \right)\sin {{0}^{\circ }}
⇒sin⁡2B=sin⁡20∘=0×0\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\times 0
⇒sin⁡2B=sin⁡20∘=0\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0
Then the value of sin⁡2A−sin⁡2B{{\sin }^{2}}A-{{\sin }^{2}}B will be
sin⁡2A−sin⁡2B=sin⁡290∘−sin⁡20∘{{\sin }^{2}}A-{{\sin }^{2}}B={{\sin }^{2}}{{90}^{\circ }}-{{\sin }^{2}}{{0}^{\circ }}
⇒sin⁡2A−sin⁡2B=1−0\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1-0
⇒sin⁡2A−sin⁡2B=1\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∘A={{45}^{\circ }} or B=45∘B={{45}^{\circ }} then sin⁡45∘\sin {{45}^{\circ }} would be sin⁡45∘=12\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} and sin⁡245∘=(sin⁡45∘)(sin⁡45∘)=(12)(12)=12.{{\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⁡245∘{{\sin }^{2}}{{45}^{\circ }} and sin⁡45∘.\sin {{45}^{\circ }}. Here it was A=90∘A={{90}^{\circ }} and sin⁡90∘=1=sin⁡290∘,\sin {{90}^{\circ }}=1={{\sin }^{2}}{{90}^{\circ }}, so that is the same here.