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If tanx=ntany ,n in R^+, then the maxi...

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)`

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`cos(x-y) = cosxcosy+sinxsiny`
`=>cos(x-y)= cosxcosy(1+tanxtany)`
`=>cos(x-y) = cosxcosy(1+(ntany)tany)`
`=>cos(x-y) = cosxcosy(1+ntan^2y)`
`=>sec(x-y) = (secxsecy)/(1+ntan^2y)`
`=>sec^2(x-y) = (sec^2xsec^2y)/(1+ntan^2y)^2`
`=>sec^2(x-y) = ((1+tan^2x)(1+tan^2y))/(1+ntan^2y)^2`
`=>sec^2(x-y) = ((1+n^2tan^2y)(1+tan^2y))/(1+ntan^2y)^2`
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