在△ABC中角ABC的对边分别为abc且cos2C=1-8b^2/a^2求1/tanA+1/tanC2、若tanB=8/15,求tanA,tanC

Question

在△ABC中 角A B C的对边分别为abc 且cos2C=1-8b^2/a^2 求1/tanA+1/tanC 2、若tanB=8/15,求tanA,tanC

Answer 1

cos2C＝1－8b²／a²
1-2(sinC)^2=1-8b²／a²
2(sinC)^2=8(sinB)^2/(sinA)^2
sinAsinC=2SinB
1.1／tanA＋1／tanC
=cosA/SinA+cosC/sinC
=(cosASinC+cosCSinA)/sinAsinC
=sin(A+C)/sinAsinC
=sinB/sinAsinC
=sinB/2sinB
=1/2