Let a solution $y=y(x)$ of the differential equation $\frac{d y}{d x} \cos x+y \sin x=1$ satisfy $y(0)=1$.

Statement-1: $y(x)=\sin \left(\frac{\pi}{4}+x\right)$

Statement-2: The integrating factor of the given differential equation is $\sec x$.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

We have,

$\frac{d y}{d x} \cos x+y \sin x=1$

$\Rightarrow \frac{d y}{d x}+y \tan x=\sec x$       ....(i)

This is a linear differential equation with integrating factor given by

Integrating factor = $e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

So, statement-2 is true.

Multiplying both sides of (i) by integrating factor $=\sec x$ and integrating w.r. to $x$, we get

$y \sec x=\tan x+C$      ........(ii)

It is given that $y=1$ when $x=0$.

∴  $1=C$

Putting $C=1$ in (ii), we get

$y \sec x=\tan x+1$

$\Rightarrow y=\sin x+\cos x=\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)$

So, statement- 1 is true and statement-2 is a correct explanation for statement- 1.