1.

tan y sec^2x dx + tan x sec^y dy +0

Answer» sec2xtanydx+sec2ytanxdy=0

On separating the variables (dividing the equation by \tan  x \tan  y)

⇒tanxsec2x​dx=−tanysec2y​dy

On integrating both sides, we get

∫tanxsec2x​dx=−∫tanysec2y​dy

Put tanx=u⇒sec2x.dx=du and tany=v⇒sec2y.dy=dv

∴∫udu​=−∫vdv​

⇒logu=−logv+logc

⇒u=vc​⇒u.v=c

∴tanx.tany=c


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