`(dy)/(dx)=(tanx)/(1+x)e^(x) secy`

Answer

`siny=(e^(x)+c)(x+1)`

Explanation

We have `(dy)/(dx)-(tany)/(1+x)=(1+x)e^(x)secy`
`therefore cosy(dy)/(dx)+(-siny)1/(1+x)=e^(x)(1+x)`…………(1)
Put `siny=y rArr cosy(dy)(dx)=(dv)/(dx)`
`therefore (1)` reduces to
`(dv)/(dx)+(-1/(1+x))v=e^(x)(1+x)`
I.F. `=e^(-1/(1+x)dx)=1/(1+x)`
`therefore` Solution is
`v.1/(x+1)=int1/(x+1)e^(x)(1+x)+c`
`rArr siny=(e^(x)+c)(x+1)`