Question

(cos66cos42+cos24cos48)/(sin^2 12-cos^2 12)

Answer 1

$\frac{cos66\cdot cos42+cos24\cdot cos48}{sin^212-cos^212}=\frac{sin24\cdot cos42+cos24\cdot sin42}{-cos24}=-\frac{sin(24+42)}{cos24}=\\\\=\frac{sin66}{cos24}=-\frac{cos24}{cos24}=-1\\\\\\P.S.\; \; cos66=cos(90-24)=sin24\\\\.\; \; \; \; \quad cos48=cos(90-42)=sin42$

Answer 2

(cos66cos42+cos24cos48)/(sin^2 12-cos^2 12)

(cos66cos42+cos24cos48)/(sin144-cos144) ≈ -0,0941943