Given that

$A+B+C=270°$

$\Rightarrow \cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C$

$=\cos 2A+\left[2\cos \left(\frac{2B+2C}{2}\right)\cos \left(\frac{2B-2C}{2}\right)\right]+4\sin A\sin B\sin C$

$=\cos 2A+2\cos (B+C)\cos (B-C)+4\sin A\sin B\sin c$

$=\cos 2A+2\cos (270°-A)\cos (B-C)+4\sin A\sin B\sin C$

$=\cos 2A-2\sin A\cos (B-C)+4\sin A\sin B\sin C$

$=1-2{\sin }^{2}A-2\sin \left(270°-\left(B+C\right)\right)\cos \left(B-C\right)+4\sin A\sin B\sin C$

$=1-2{\sin }^{2}A-2\sin A\cos (B-C)+4\sin A\sin B\sin C$

$=1-2\sin A\left(\sin A+\cos \left(B-C\right)\right)+4\sin A\sin B\sin C$

$=1-2\sin A\left(-\cos (B+C)+\cos (B-C)\right)+4\sin A\sin B\sin C$

$=1-2\sin A\left(2\sin \left(\frac{B+C+B-C}{2}\right)\sin \left(\frac{(B+C)-(B-C)}{2}\right)\right)+4\sin A\sin B\sin C$

$=1-2 \sin A\left(2\sin B\sin C\right)+4\sin A\sin B\sin C$

$=1$