`sec^(2) y tan x dy + sec^(2) x tan y dx = 0 , " when " x =y = pi/4`

Explanation

The given D.E .is ` sec^(2)y tan x dy + sec^(2) x tan y dx = 0`
`(sec^(2)y)/(tany) dy + (sec^(2)x)/tanx dx =0`
`int (sec^(2)y)/(tan y) dy + int (sec^(2)x)/(tanx) dx=c_(1)`
Each integral is of the type
` int (f'(x))/(f(x) dx = log|f(x)|+c`
`log |tany| + log | tan x| = log c, " where" c_(1) = log c`
log | tan x tan y| = log c
This is the general solution ,.
When ` x=y = pi/4,` we get s,
` tan pi/4. tan pi/4 " " therefore c=1`
the particular solution is tan x tan y=1