Answer
\[2\cos x\cos y=1\]
Work Step by Step
$\frac{dy}{dx}=1-\frac{\sin(x+y)}{\sin y\cos x}$ ___(1)
$\frac{dy}{dx}=1-\frac{\sin x\cos y+\sin y\cos x}{\sin y\cos x}$
$\frac{dy}{dx}=1-\frac{\tan x}{\tan y}+1=\frac{-\tan x}{\tan y}$
Separating variablea,
$-\tan y\;dy=\tan x \;dx$
Integrating,
$-\int\tan y\;dy=\int\tan x \;dx+\ln C$
$\ln C$ is constant of integration
$\ln|\cos y|=-\ln|\cos x|+\ln|C|$
$\ln|\cos x\cos y|=\ln|C|$
$\cos x\cos y=C$ ___(2)
Using initial condition $y\left(\frac{π}{4}\right)=\frac{π}{4}$
$\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}=\frac{1}{2}=C$
(2) becomes
\[2\cos x\cos y=1\]
Hence solution of (1) is $2\cos x\cos y=1$.
