The given differential equation is:
ex tan y dx + (1−ex)sec2y dy = 0
⇒ ex tan y dx = - (1−ex)secy dy
⇒ ex tan y dx = (ex−1)secy dy
⇒ ex−ex−1 dx =

sec2y

tany

dy
On integrating on both sides, we get
∫ ex−ex−1 dx = ∫

sec2y

tany

dy ... (1)
Let I1 = ∫

sec2y

tany

dy
Put tany = t
⇒ sec2 y dy = t
∴ ∫

sec2y

tany

dy = ∫

dt

t

= log |t| = log tan y ... (2)
Let I2 = ∫

ex

ex−1

dx
Put ex - 1 = u
∴ ex dx = du
∫

ex

ex−1

dx = ∫

du

u

= log u
= log (ex−1) ... (3)
From i , ii and iii , we get
log tan y = log (ex - 1) + log C
⇒ log tan y = log C (ex - 1)
⇒ tan y = C (ex - 1)
The solution of the given differential equation is tan y = C (ex - 1).