`a-b-a^2+b^2`

Prove |[1+a^2-b^2, 2 a b, -2 b],[ 2 a b, 1-a^2+b^2, 2 a],[ 2 b, -2 a, 1-a^2-b^2]|=(1+a^2+b^2)^3

Simplify: a^2b(a-b^2)+a b^2(4a b-2a^2)-a^3b(1-2b)

Simplify: 4a b(a-b)-6a^2(b-b^2)-3b^2(2a^2-a)+2a b(b-a)

By using properties of determinants. Show that: |[1+a^2-b^2, 2a b,-2b],[2a b,1-a^2+b^2, 2a],[2b,-2a,1-a^2-b^2]|=(1+a^2+b^2)^3

Show that: |1+a^2-b^2 2a b-2b2a b1-a^2+b^2 2a2b-2a1-a^2-b^2|=(1+a^2+b^2)^3

If tantheta=a/b , then (asintheta+bcostheta)/(asintheta-bcostheta) is equal to (a) (a^2+b^2)/(a^2-b^2) (b) (a^2-b^2)/(a^2+b^2) (c) (a+b)/(a-b) (d) (a-b)/(a+b)

What is the value of {((a+b)^(2) - (a^(2) +b^(2)))/((a+b)^(2) - (a-b)^(2))}?

If sin theta + cos theta =a, and sin theta - cos theta =b, then ..... (A) a^2 + b^2 =1 (B) a^2 - b^2 = 1 ( C) a^2 + b^2 =2 (D) a^2 - b^2 =2