tan x =n tan y n belongs to R+ then maximum value of sec^2(x-y)=(n+1)^2/4n

 

Asked by aahnik.mohanty | 29th May, 2019, 08:04: PM

Expert Answer:

cos (x - y) = cosx cosy + sinx siny
                = cosx cosy (1 + tanx tany)
                = cosx cosy (1 + ntan2 y)
sec2 (x - y) = 1/cos2 (x - y)
                 = sec2 x sec2 y/(1 + ntan2 y)
                 = (1 + tan2 x)(1 + tan2 y)/ (1 + ntan2 y)2
                 = (1 + ntan2 y)(1 + tan2 y)/(1 + ntan2 y)2
Solving we get
sec2 (x - y) = [1 + (n - 1)2 tan2 y]/(1 + ntan2 y)2
1 + ntan2 y ≥ ntan2 y
[(1 + ntan2 y)/2]2 ≥ ntan2 y                    AM ≥ GM
tan2 y/(1 + ntan2 y)2 ≤ 1/4n
sec2 (x - y) ≤ 1 + (n - 1)2/4n = (n + 1)2/4n

Answered by Sneha shidid | 30th May, 2019, 10:13: AM