How do you solve \[{{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)+2\sin A\cos B\sin \left( B-A \right)\] ?
A. \[{{\sin }^{2}}A\]
B. \[{{\sin }^{2}}B\]
C. \[{{\cos }^{2}}A\]
D. \[{{\cos }^{2}}B\]

Hint: In the given question, we have been asked to solve the trigonometric expression i.e. \[{{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)+2\sin A\cos B\sin \left( B-A \right)\] . In order to solve this, we will expand or simplify the given trigonometric expression by using the trigonometric identity that i.e. \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] . Later by using the distributive property of multiplication we will simplify the given expression and solve the terms. In this way we will get the required answer.
We can use the trigonometric relations to write the given expressions in different trigonometric functions;
I. \[\cot x=\dfrac{1}{\tan x}\]
II. \[\tan x=\dfrac{\sin x}{\cos x}\]
III. \[\cot x=\dfrac{\cos x}{\sin x}\]
IV. \[\sec x=\dfrac{1}{\cos x}\]
V. \[co\sec x=\dfrac{1}{\sin x}\]

Complete step by step solution:
We have given that,
 \[\Rightarrow {{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)+2\sin A\cos B\sin \left( B-A \right)\]
Now,
Using the trigonometric identity i.e. \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]
Thus,
Substitute the value of \[{{\sin }^{2}}\left( A-B \right)=\left( \sin A\cos B-\cos A\sin B \right)\left( \sin A\cos B-\cos A\sin B \right)\] in the above expression, we will obtain
 \[\Rightarrow {{\sin }^{2}}A+\left( \sin A\cos B-\cos A\sin B \right)\left( \sin A\cos B-\cos A\sin B \right)+2\sin A\cos B\left( \sin B\cos A-\cos B\sin A \right)\]
Using the distributive property of multiplication i.e. \[\left( a+b \right)\times \left( c+d \right)=ac+ad+bc+bd\]
Simplifying the above equation,
\[\Rightarrow {{\sin }^{2}}A+\left( {{\sin }^{2}}A{{\cos }^{2}}B-\sin A\sin B\cos A\cos B-\sin A\sin B\cos A\cos B+{{\cos }^{2}}A{{\sin }^{2}}B \right)+2\sin A\sin B\cos A\cos B-2{{\cos }^{2}}B{{\sin }^{2}}A\]
Combining the like terms, we will get
 \[\Rightarrow {{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B-2\sin A\sin B\cos A\cos B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\sin B\cos A\cos B-2{{\sin }^{2}}A{{\cos }^{2}}B\]
Again
Combining the like terms, we will get
 \[\Rightarrow {{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B-2{{\sin }^{2}}A{{\cos }^{2}}B\]
We will obtain,
 \[\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B\]
Taking \[{{\sin }^{2}}A\] common term,
 \[\Rightarrow {{\sin }^{2}}A\left( 1-{{\cos }^{2}}B+{{\cot }^{2}}A{{\sin }^{2}}B \right)\]
 \[\Rightarrow {{\sin }^{2}}A{{\sin }^{2}}B\left( 1+{{\cot }^{2}}A \right)\]
As we know that, \[\left( 1+{{\cot }^{2}}A \right)=\cos e{{c}^{2}}A\]
Substituting the value, we will get
 \[\Rightarrow {{\sin }^{2}}A{{\sin }^{2}}B\left( \cos e{{c}^{2}}A \right)\]
As we know that the relation between the sin and cosec trigonometric function is \[\cos ec\theta =\dfrac{1}{\sin \theta }\] ,
Therefore,
 \[\Rightarrow {{\sin }^{2}}A{{\sin }^{2}}B\left( \dfrac{1}{si{{n}^{2}}A} \right)\]
Cancelling out the common terms, we get
 \[\Rightarrow {{\sin }^{2}}B\]
Therefore,
 \[\Rightarrow {{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)+2\sin A\cos B\sin \left( B-A \right)={{\sin }^{2}}B\]
Hence, the option (b) is the correct answer.
So, the correct answer is “Option b”.

Note: To solve any given trigonometric expression, you just need to remember all the basic identities of trigonometric functions and simplify the equation or the expression by using them. Since all trigonometric functions are inter-relatable thus it may be generally possible to solve the given trigonometric expression in multiple ways to arrive at the solution.