Bila tan A = -3/4, nilai 3 cos A + 2 sin A per cos A - 4 sin A = ....
fufutiensov2brh
Tan A= y/x= -3/4 (bisa di kuadran II atau IV). Jika asumsiny nilai negatifnya itu y dan x positif maka berada di kuadran IV. Dgn demikian r^2 = x^2 + y ^2 r^2 = (-3)^2 + (4)^2 r^2= 9 + 16 r^2 = 25 r = 5
cos A= x/r = 4/5 sin A = y/r= -3/5
Nilai (3 cos A + 2 sin A) / (cos A - 4 sin A) = (3. 4/5 + 2. -3/5) / (4/5 - 4. -3/5) = (12/5 + -6/5) / (4/5 + 12/5) = 6/5 : 16/5 = 6/5 × 5/16 = 6/16 = 3/8
Dgn demikian
r^2 = x^2 + y ^2
r^2 = (-3)^2 + (4)^2
r^2= 9 + 16
r^2 = 25
r = 5
cos A= x/r = 4/5
sin A = y/r= -3/5
Nilai (3 cos A + 2 sin A) / (cos A - 4 sin A) = (3. 4/5 + 2. -3/5) / (4/5 - 4. -3/5)
= (12/5 + -6/5) / (4/5 + 12/5)
= 6/5 : 16/5
= 6/5 × 5/16
= 6/16
= 3/8
Semoga benar, maaf jika ada kekeliruan.