Question

If ${\sin }^{4}x+{\cos }^{4}y+2=4\sin x\cos y(0\le x,y\le \frac{\pi }{2})$, then the value of $\sin x+\cos y$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

Answer: (C)

$2$

The equation is ${\sin }^{4}x+1+{\cos }^{4}y+1-4\sin xcosy=0$
$({\sin }^{2}x-1{)}^2+({\cos }^{2}y-1{)}^2+2(\sin x-cosy{)}^2=0$
$\Rightarrow {\sin }^{2}x=1,{\cos }^{2}y=1,\sin x=\cos y$
$\sin x=1,\cos y=1,\sin x=\cos y$
Because $0\le x,y\le \frac{\pi }{2},\sin x=1\Rightarrow x=\frac{\pi }{2}$
$\cos y=1\Rightarrow y=0$
$\Rightarrow \sin x+\cos y=2$