Solve the following differential equation: `tanydx+sec^2ytanx dy=0`

Explanation

`tanydx+sec^2ytanx dy=0`


on seprating variables, we get



`-frac{d x}{tan x}=frac{sec ^{2} y}{tan y} d y`


`-cot (x) d x=sec ^{2} y cot (y) d y`


Integrating both sides we get,


`-int cot (x) d x=int sec ^{2} y cot (y) d y `


`-int cot (x) d x=int frac{1}{sin y cos y} d y`


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