tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1 求cosa*|tana|

tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1 求cosa*|tana|

tan(a-4π)=-tanacos(π+a)=-cosasin^2(a+3π)=sin^2atan(3π+a)=-tanacos^2(5π/2+a)=sin^2a方程左邊=-cosa∴cosa=1/2|tana|=根號3所以答案=二分之根號三這位朋友打字錯誤太多.5/2寫成2/5,1/2寫成2/1.π寫成t.最後...