If x is the measure of an acute angle , simplify the expression E=sin^2x*cos^2x*[1/square root[1-sin^2x*(1+cos^2x)]+1/square...

We'll start with the denominator of the first fraction and
we'll open the brackets of radicand.

sqrt[1-(sin x)^2-(sin
x)^2*(cosx)^2]

From Pythagorean theorem,we'll
get:

1-(sin
x)^2=(cosx)^2

sqrt[1-(sin x)^2-(sin
x)^2*(cosx)^2]=sqrt[(cosx)^2-(sin
x)^2*(cosx)^2]

sqrt[(cosx)^2-(sin
x)^2*(cosx)^2]=sqrt(cosx)^2*[1-(sin
x)^2]

sqrt(cosx)^2*[1-(sin x)^2] =
sqrt[(cosx)^2*(cosx)^2]=(cosx)^2

The first fraction inside
brackets will become:

1/sqrt[1-(sin
x)^2*(1+(cosx)^2)]=1/(cosx)^2 (1)

We'll do the
same steps for the 2nd fraction.

sqrt[1-(cos x)^2-(sin
x)^2*(cos x)^2]

From Pythagorean theorem,we'll
get:

1-(cos x)^2 = (sin
x)^2

sqrt[1-(cos x)^2-(sin x)^2*(cos x)^2]=sqrt[(sin
x)^2-(sin x)^2*(cos x)^2]

sqrt[(sin x)^2-(sin x)^2*(cos
x)^2]=sqrt(sin x)^2*[1-(cos x)^2]

sqrt(sin x)^2*[1-(cos
x)^2]=sqrt[(sin x)^2*(sin x)^2]=(sin x)^2

The second
fraction will
become:

1/sqrt[1-(cosx)^2*(1+(sinx)^2)]=1/(sin x)^2
(2)

We'll re-write the expression, replacing the
fractions by (1) and (2):

E = (sin
x)^2*(cosx)^2*[1/(cosx)^2 + 1/(sin x)^2]

E = (sin
x)^2*(cosx)^2*[(sin x)^2+(cosx)^2]/(cosx)^2*(sin x)^2

But
(sin x)^2+(cosx)^2=1

E = (sin x)^2*(cosx)^2/(cosx)^2*(sin
x)^2

E = 1

The
requested simplified expression is: E =
1.