1.

The solution of differential equation `cos x (dy)/(dx)+y sin x =1`A. `tanx+tany=k`B. `tanx-tany=k`C. `(tanx)/(tany)=k`D. `tanx.tany=k`

Answer» Given that, `" "tanysec^(2)xdx+tanxsec^(2)ydy=0`
`rArr" "tansec^(2)xdx=-tanxsec^(2)ydy`
`rArr" "(sec^(2)x)/(tanx)dx=(-sec^(2)y)/(tany)dy" "`...(i)
On integrating both sides, we have
`" "int(sec^(2)x)/(tanx)dx=-int(sec^(2)y)/(tany)dy`
Put `tanx=t` in LHS integral, we get
`" "sec^(2)xdx=dtrArrsec^(2)xdx=dt`
and `" "tany=u` in RHS integral, we get
`" "sec^(2)ydy=du`
On substituting these values in Eq. (i), we get
`" "int(dt)/(t)=-int(du)/(u)`
`" "logt=-logu+logk`
`rArr" "log(t*u)=logk`
`rArr" "log(tanxtany)=logk `
`rArr" "tanxtany=k`


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