Solution of cos x dy/dx y sin x =1 is y sec x = tan x + c.

Explanation:
Given differential equation is cos x dy/dx y sin x =1.
Let us divide both sides by cos x sin x, we get
dy/dx = 1/(cos x sin x)
Integrating both sides, we get
∫dy = ∫dx/(cos x sin x)
Using the formula, ∫dx/(cos x sin x) = (1/2) ln|tan(x/2)| + c1
where c1 is the constant of integration.
Therefore, y = (1/2) ln|tan(x/2)| + c2
where c2 is another constant of integration.
Simplifying the equation, we get
y = ln|√(tan x)| + c2
y = ln|sec x tan x| + c2
y = ln|sec x| + ln|tan x| + c2
y = ln|sec x| + ln|sin(x + π/2)| + c2
y = ln|sec x| + ln|cos (π/2 – x)| + c2
y = ln|sec x| – ln|sin x| + c2
y = ln|sec x/sin x| + c2
y = ln|cosec x – cot x| + c2
Using the identity, sec x = 1/cos x = √(1 – sin^2 x)/sin x = √(cos^2 x – 1)/cos x,
we get
y = ln|√[(cos^2 x – 1)/cos^2 x]/sin x| + c2
y = ln|√[(1 – tan^2 x)/cos^2 x]/sin x| + c2
y = ln|√[1/cos^2 x – tan^2 x/cos^2 x]/sin x| + c2
y = ln|√[sec^2 x – tan^2 x]/sin x| + c2
y = ln|√(sec x + tan x)(sec x – tan x)/sin x| + c2
y = ln|√(sec x + tan x)/sin x| + ln|√(sec x – tan x)| + c2
y = ln|sec x + tan x|/2 + ln|sec x – tan x|/2 + c2
y = ln|sec x + tan x| – ln|cos x| + c2
y = ln|sec x + tan x| – ln|sec x| + c2
y = ln|sec x (sec x + tan x)| – ln|sec x| + c2
y = ln|sec x tan x + sec x| – ln|sec x| + c2
y = ln|sec x tan x + 1| + ln|1/sec x| + c2
y = ln|sec x tan x + 1| – ln|cos x| + c2
y = ln|sec x tan x + 1| – ln|sin (π/2 – x)| + c2
y = ln|sec x tan x +