Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requires going from:
Expression 1: $$-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2$$
to:
Expression 2: $$(a+b-c)(a-b+c)(a+b+c)(-a+b+c)$$
I can see that these two expressions are equal by working in reverse and multiplying out the second expression to get the first but how does one go from Exp.1 to Exp.2?

$$\begin{align}&-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2\\&=-a^4+(2b^2+2c^2)a^2-b^4+2b^2c^2-c^4\\&=-a^4+2(b^2+c^2)a^2-(b^2-c^2)^2\\&=\color{red}{-a^4+2(b^2+c^2)a^2-(b^2+c^2)^2}+(b^2+c^2)^2-(b^2-c^2)^2\\&=\color{red}{-(a^2-(b^2+c^2))^2}+4b^2c^2\\&=(2bc)^2-(a^2-b^2-c^2)^2\end{align}$$
Now use $A^2-B^2=(A+B)(A-B)$.
