Jika 3cosB = 5sinB, maka... (B = Simbol Beta) 5sinB - 2sec^3 B + 2cosB -------------------------------------- = ??? 5sinB + 2sec^3 B - 2co

3 cos B = 5 sin B
5 sin B = 3 cos B
(sin B)/(cos B) = 3/5
Tan B = 3/5 = de/sa => mi = √(5^2 + 3^2) = √(25 + 9) = √34

Sin B = de/mi = 3/√34 => 5 sin B = 15/√34
Cos B = sa/mi = 5/√34 => 2 cos B = 10/√34
Sec B = mi/sa = √34/5 => sec^3 B = (34√34)/125

(5 sin B - 2 cos^3 B + 2 cos B)/(5 sin B + 2 sec^3 B - 2 cos B)
= (15/√34 - 2(34√34)/125 + 10/√34)/(15/√34 + 2(34√34)/125 - 10/√34)
= (15 - 2312/125 + 10)/(15 + 2312/125 - 10)
= (25 - 2312/125) / (5 + 2312/125) . 125/125
= (3125 - 2312) / (625 + 2312)
= 813/2937
= 271/979