If `tanx =n tany , n in R^+` then the maximum value of `sec^2(x-y)` is

If tan x = n tan y, nin R ' then the maximum value of sec^(2)(x-y) is equal to

If tanx=ntany ,n in R^+, then the maximum value of sec^2(x-y) is equal to (a) ((n+1)^2)/(2n) (b) ((n+1)^2)/n (c) ((n+1)^2)/2 (d) ((n+1)^2)/(4n)

If tanx=ntany ,n in R^+, then the maximum value of sec^2(x-y) is equal to (a) ((n+1)^2)/(2n) (b) ((n+1)^2)/n (c) ((n+1)^2)/2 (d) ((n+1)^2)/(4n)

If tanx=ntany ,n in R^+, then the maximum value of sec^2(x-y) is equal to (a) ((n+1)^2)/(2n) (b) ((n+1)^2)/n (c) ((n+1)^2)/2 (d) ((n+1)^2)/(4n)

If tan x=n tan y,n in R^(+), then the maximum value of sec^(2)(x-y) is equal to (a) ((n+1)^(2))/(2n) (b) ((n+1)^(2))/(n)( c) ((n+1)^(2))/(2) (d) ((n+1)^(2))/(4n)

If x, y, k, m, n are positive and x + y = k, then find the maximum value of x^(m)y^(n) .

If tanx+tany=25 and cotx+coty=30 then the value of tan(x+y) is

If tanx=1/7,tany=1/3 ,then x+y is:

If x+y =pi/4 and tanx + tany=1 , then