Question

Find the value of ${\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B)$.

Hint: We will try to simplify the given expression by using the formula: $\sin (A - B) = \sin A\cos B - \cos A\sin B$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$. Doing all this and simplifying, we will get the desired value.

We have with us the expression in the question which is: ${\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B)$.
Let us put in the value of $\sin (A - B) = \sin A\cos B - \cos A\sin B$.
We will now get:-
$\Rightarrow {\sin ^2}B - {\sin ^2}A - {(\sin A\cos B - \cos A\sin B)^2} + 2\sin A\cos B(\sin A\cos B - \cos A\sin B)$
Now, we will apply ${(a - b)^2} = {a^2} + {b^2} - 2ab$ on ${(\sin A\cos B - \cos A\sin B)^2}$by taking $a = \sin A\cos B$ and $b = \cos A\sin B$ to get ${(\sin A\cos B - \cos A\sin B)^2} = {(\sin A\cos B)^2} + {(\cos A\sin B)^2} - 2.\sin A\cos B.\cos A\sin B$
So, we get:
$\Rightarrow {\sin ^2}B - {\sin ^2}A - {(\sin A\cos B)^2} - {(\cos A\sin B)^2} + 2.\sin A\cos B.\cos A\sin B + \\ 2\sin A\cos B(\sin A\cos B - \cos A\sin B) \\$
Simplifying the expression further to get the following expression:
$\Rightarrow {\sin ^2}B - {\sin ^2}A - {\sin ^2}A{\cos ^2}B - {\cos ^2}A{\sin ^2}B + 2\sin A\sin B\cos A\cos B + \\ 2{\sin ^2}A{\cos ^2}B - 2\sin A\sin B\cos A\cos B \\$
Simplifying the expression further by clubbing the like terms to get the following expression:
$\Rightarrow {\sin ^2}B - {\sin ^2}A + {\sin ^2}A{\cos ^2}B - {\cos ^2}A{\sin ^2}B$
We can rewrite it as:-
$\Rightarrow {\sin ^2}B(1 - {\cos ^2}A) - {\sin ^2}A(1 - {\cos ^2}B)$
Now, we will apply the formula: ${\sin ^2}\theta + {\cos ^2}\theta = 1$ in here to get: ${\sin ^2}{\rm B} = 1 - {\cos ^2}{\rm B}$ and ${\sin ^2}{\rm A} = 1 - {\cos ^2}{\rm A}$.
$\Rightarrow {\sin ^2}B {\sin ^2}A - {\sin ^2}A {\sin ^2}B$
$\Rightarrow 0$
Hence, the value of the expression ${\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B)$ is 0.