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

Expert Answers
justaguide eNotes educator| Certified Educator

We have to simplify: E=sin^2x*cos^2x*[1/square root[1-sin^2x*(1+cos^2x)]+1/square root[1-cos^2x*(1+sin^2x)]]

=> sin^2x*cos^2x*[1/square root[1-(1-cos^2x)*(1+cos^2x)]+1/square root[1-(1-sin^2x)*(1+sin^2x)]]

=> sin^2x*cos^2x*[1/square root[1-(1-(cos^2x)^2)]+1/square root[1-(1-(sin^2x)^2]]

=> sin^2x*cos^2x*[1/square root[(cos^2x)^2)]+1/square root[(sin^2x)^2]]

=> sin^2x*cos^2x*[1/cos^2x +1/(sin^2x)]

=> sin^2x*cos^2x*[((sin^2x) + (cos^2x))/(cos^2x)*(sin^2x)]


The simplified expression is equal to 1.

giorgiana1976 | Student

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.