Question

If $\frac{dy}{dx}=tan^{2}\left(x+y\right)$, then $sin2\left(x+y\right)=$

• A. $x - y + c$
• B. $2(x - y) + c$
• C. $x + y + c$
• D. $2(x + y) + c$

Answer 1

Answer: (B)

Substitute $x+y=z$
$ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$
So, the given equation becomes
$\frac{dz}{dx}=sec^{2}\,z$
$\Rightarrow dx=cos^{2}\,z\,dz=\frac{1+cos\,2z}{2}dz$
$\Rightarrow x=\frac{1}{2}\left(z+\frac{sin\,2z}{2}\right)+c_{1}$
$\Rightarrow 2x=x+y+\frac{sin\,2\left(x+y\right)}{2}+c_{2}$
$\Rightarrow 2\left(x-y\right)=sin\,2\left(x+y\right)-c$
$\Rightarrow sin2\left(x+y\right)=2\left(x-y\right)+c$