Упростите выражение под б и г

Решите задачу:

$1)\; \; \frac{cos6+cos12+cos36+cos42}{sin87\cdot cos15\cdot cos24} = \frac{(cos6+cos42)+(cos12+cos36)}{sin(90-3)\cdot cos15\cos24} =\\\\= \frac{2\cdot cos24\cdot cos18+2\cdot cos24\cdot cos12}{cos3\cdot cos15\cdot cos24} = \frac{2\cdot cos24\cdot (cos18+cos12)}{cos3\cdot cos15\cdot cos24} =\\\\= \frac{2\cdot 2\cdot cos15\cdot cos3}{cos3\cdot cos15}=4$

$2)\; \; \frac{sin48-sin60-sin72+sin84}{4cos84\cdot sin12\cdot sin66} = \frac{(sin48+sin84)-(sin60+sin72)}{4\cdot cos(90-6)\cdot sin12\cdot sin66} =\\\\= \frac{2\cdot sin66\cdot cos18-2\cdot sin66\cdot cos \, 6}{4\cdot sin6\cdot sin12\cdot sin66} = \frac{2\cdot sin66\cdot (cos18-cos\, 6)}{4\cdot sin6\cdot sin12\cdot sin66} =\\\\= \frac{-2\cdot sin12\cdot sin6}{2\cdot sin6\cdot sin12} =-1$