If sin4x+cos4y+2=4sinxcosy,0≤x,y≤π/2 then sinx+cosy=

The given equation can be written as sin4x+cos4y+2−4 sin xcosy=0⇒sin2x−12+cos2y−12+2sin2x+2cos2y−4sinxcosy=0⇒sin2x−12+cos2y−12+2sinx−cosy2=0which is true if sin2x=1,cos2y=1 and sinx=cosy,so sin x+cos y=2 as 0≤x,y≤π/2.

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If $\sin^{4} x + \cos^{4} y + 2 = 4$ $\sin x \cos y, 0 \leq x, y \leq π / 2$ then $\sin x + \cos y =$

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Detailed Solution

The given equation can be written as
$\sin^{4} x + \cos^{4} y + 2 - 4 \sin x \cos y = 0$
$\Rightarrow {(\sin^{2} x - 1)}^{2} + {(\cos^{2} y - 1)}^{2} + 2 \sin^{2} x + 2 \cos^{2} y - 4 \sin x \cos y = 0$
$\Rightarrow {(\sin^{2} x - 1)}^{2} + {(\cos^{2} y - 1)}^{2} + 2 {(\sin x - \cos y)}^{2} = 0$
which is true if $\sin^{2} x = 1, \cos^{2} y = 1$ and $\sin x = \cos y,$
so $\sin x + \cos y = 2$ as $0 \leq x, y \leq π / 2$ .