help me.. with these dumb doubts

i have edited it manmay...

and in 3rd question.. how can it be e?

FOR EQN. OF THE FORM dydx+Ky = c
I.F = e^∫kdx sorry mistook. answer is e^x.
deleted that post. WAS A TYPO.

2) I.F = cot x
If correct i will post full soln.

P(A∩B)=P(A) then

P(B/A)=P(A∩B)P(A)=1

2) \frac{dy}{dx}-\frac{sec^{2}x}{tanx}y=\frac{sinx}{tanx}
= \frac{dy}{dx}-\frac{1+tan^{2}x}{tanx}y=cosx
= \frac{dy}{dx}-\left( \frac{1}{tanx}+{tanx}\right)y=cosx

comparing with general form \Rightarrow \frac{dy}{dx}+Ky = C where I.F = e^{∫K dx}

we get I.F. = e^{-\int (cotx+tanx)dx}
= e^{-\int \frac{cosx}{sinx}dx-\int \frac{sinx}{cos x}dx}
= e^{-ln\left|sinx \right|+ln\left|cosx \right|}
= e^{ln\left|cosecx \right|+ln\left|cosx \right|}
= e^{ln (cosecx .cosx)}\Rightarrow e^{log_{e}cotx}\Rightarrow cotx

HENCE I.F. = cot x

what is IF
and how is
comparing with general form where I.F = e∫K dx