Question

ptove that : cos a + 2 cos 3a +cos 5a/cos 3a+2cos5a+cos7a=cos3asec5a​

Answer 1

Answer:

(cos3A+2cos5A+cos7A)/( cosA+2cos3A+cos5A) = 2cos 5A [cos 2A + 1] / 2cos 3A [cos 2A + 1] = cos 5A / cos 3A = [cos 3A

Step-by-step explanation:

(cos3A+2cos5A+cos7A)/(cosA+2cos3A+cos5A) = 2cos 5A [cos 2A + 1] / 2cos 3A [cos 2A