Question

Solve :- $\frac{dy}{dx}=\tan \left(x+y\right)$

Answer 1

$\frac{dy}{dx}=\tan (x+y)$

$\frac{dy}{dx}=\frac{\sin (x+y)}{\cos (x+y)}$

Let $x+y=v$

$1+\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dv}{dx}-1=\frac{\sin v}{\cos v}$

$\Rightarrow dv=\frac{\sin v+\cos v}{\cos v}dx$

$\Rightarrow \frac{\cos v}{\sin v+\cos v}dv=dx$

$=\frac{1}{2}\int \frac{(\cos v+\sin v)}{\sin v+\cos v}+\frac{(\cos v-\sin v)}{\sin v+\cos v}dv=\int dx$

$=\frac{1}{2}\int dv+\frac{1}{2}\int \frac{\cos v-\sin v}{\sin v+\cos v}dv=\int dx$

$=\frac{1}{2}v+\frac{1}{2}log|\sin v+\cos v|=x+c$

$=\frac{1}{2}(x+y)+\frac{1}{2}log|\sin (x+y)+\cos (x+y)|=x+c$

$=y-x+log|\sin (x+y)+\cos (x+y)|=c$.