# Solution of the differential equation t any sec2x dx + tanx sec2 ydy = 0 is:A. tanx + tany = kB. tanx – tan y = kC. D. tanx . tany = k

The given differential equation is

tanysec2xdx + tanxsec2ydy = 0

Divide throughout by tanxtany Integrate Put tanx = t hence that is sec2xdx = dt

Put tany = z hence that is sec2ydy = dt log t + log z + c = 0

Resubstitute t and z

log(tan x) + log(tan y) + c = 0

Using log a + log b = log ab

log(tan x tan y) = -c

tan x tan y = e-c

e is a constant -c is a constant hence e-c is a constant, let it be denoted as k hence k = e-c

tan x tan y = k

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