Question

If $\frac{dy}{dx}=sin\left(x+y\right)+cos\left(x+y\right)$, $y\left(0\right)=0$, then $tan\frac{x+y}{2}=$

• A. $e^x-1$
• B. $\frac{e^x-1}{2}$
• C. $2(e^x-1)$
• D. $1-e^x$

Answer 1

Answer: (A)

Substitute $x+y=z$
$\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$
So, given equation becomes
$\frac{dz}{dx}=1+sinz+cosz$
$\Rightarrow \int dx=\int \frac{dz}{1+sinz+cosz}$, Substitute $tan\frac{z}{2}=t$
$=\int \frac{dt}{t+1}=ln\left(t+1\right)+c$
$\Rightarrow x=ln\left(1+tan\frac{x+y}{2}\right)+c$
$x=0$,
$y=0$
$\Rightarrow c=0$
$\Rightarrow e^x-1=tan\left(\frac{x+y}{2}\right)$.