Question

If `tanx =n tany , n in R^+` then the maximum value of `sec^2(x-y)` is
A. `((n+1)^(2))/(2n)`
B. `((n+1)^(2))/(n)`
C. `((n+1)^(2))/(2)`
D. `((n+1)^(2))/(4n)`

Answer 1

Correct Answer - D
`tan x=n tan y, cos(x-y)=cos x cos y+sin(xsiny)`
`rArr cos(x-y)=cos x cos y(1+tan x tan y)`
`=cos x cos y(1+n tan^(2)y)`
`rArr sec^(2)(x-y)=(sec^(2)xsec^(2)y)/((1+n tan ^(2)y)^(2))`.
`=((1+tan^(2)x)(1+tan^(2)y))/((1+n tan^(2)y)^(2))`
`=((1+n^(2)tan^(2)y)(1+tan^(2)y))/((1+tan^(2)y)^(2))`
`=1+((n-1)^(2)tan^(2)y)/((1+ntan^(2)y)^(2))`.
Now `((1+n tan^(2)y)/(2))^(2)ge n tan^(2)y`
or `(tan^(2)y)/((1+n tan^(2)y)^(2)le(1)/(4n)`
`rArr sec^(2)(x-y)le1+((n-1)^(2))/(4n)=((n+1)^(2))/(4n)`