Question

Помогите решить 50б!
$\frac{11cos \alpha +12sin \alpha }{13cos \alpha -14sin \alpha } ; tg \alpha = \frac{15}{16}$

Answer 1

$\frac{11cos \alpha +12sin \alpha }{13cos \alpha -14sin \alpha } = \frac{ \frac{11cos \alpha +12sin \alpha }{cos \alpha } }{ \frac{13cos \alpha -14sin \alpha }{cos \alpha } }= \frac{11+12 \frac{sin \alpha }{cos \alpha } }{13-14 \frac{sin \alpha }{cos \alpha } } =$$\frac{11+12tg \alpha }{13-14 tg \alpha } = \frac{11+12* \frac{15 }{16} }{13-14* \frac{15 }{16} } =\frac{11*8+6*15 }{13*8-7*15 } = \frac{88+90}{104-105} =-178$