Find the exact value of the expression by using appropriate identities. Do not use a calculator. $\sin {72}^{\circ }\cos {42}^{\circ }-\cos {72}^{\circ }\sin {42}^{\circ }$
$\sin {72}^{\circ }\cos {42}^{\circ }-\cos {72}^{\circ }\sin {42}^{\circ }=$ (Type a fraction )

Step 1 ：Let's denote $a=\sin {72}^{\circ }$ and $b=\cos {42}^{\circ }$. Then we have $\sin {72}^{\circ }\cos {42}^{\circ }-\cos {72}^{\circ }\sin {42}^{\circ }=a\cdot b-(1-a^2)\cdot (1-b^2)$.

Step 2 ：Expanding the right side, we get $a\cdot b-1+a^2+b^2-a^2\cdot b^2$.

Step 3 ：Rearranging the terms, we get $a^2+b^2+a\cdot b-a^2\cdot b^2-1$.

Step 4 ：Notice that $a^2+b^2+2\cdot a\cdot b=(a+b{)}^2$. So we can rewrite the expression as $(a+b{)}^2-2\cdot a\cdot b-1$.

Step 5 ：From the solution of QuestionA, we know that $a-b=\frac{1}{2}$. So $a+b=1-(a-b)=1-\frac{1}{2}=\frac{1}{2}$.

Step 6 ：Substituting $a+b=\frac{1}{2}$ into the expression, we get ${\left(\frac{1}{2}\right)}^2-2\cdot a\cdot b-1=\frac{1}{4}-2\cdot a\cdot b-1$.

Step 7 ：Solving for $2\cdot a\cdot b$, we get $2\cdot a\cdot b=\frac{1}{4}-1=-\frac{3}{4}$.

Step 8 ：So the original expression $\sin {72}^{\circ }\cos {42}^{\circ }-\cos {72}^{\circ }\sin {42}^{\circ }$ equals to $-\frac{3}{4}$.

Step 9 ：Finally, we check the result. Since $-1\le \sin x,\cos x\le 1$, the result $-\frac{3}{4}$ is within the range, so it is correct.

Step 10 ：Thus, the exact value of the expression $\sin {72}^{\circ }\cos {42}^{\circ }-\cos {72}^{\circ }\sin {42}^{\circ }$ is $\boxed{{\displaystyle -\frac{3}{4}}}$.