To prove 1/a(a+1) = 1/a -1/(a+1).

We start from the right side and show it is equal to the expression on the left.

RHS = 1/a -1/(a+1) .

The LCM of the denominators is a(a+1).

Therefore we convert each of the fractions 1/a and 1/(a+1) into equivalent fraction with LCM as the denominator.

So the first term 1/a = 1*(a+1)/a*(a+1) = (a+1)/a(a+1).

The second term 1/(a+1) = 1*a/(a+1)*a = a/(a(a+1).

Therefore 1/a -1/(a+1) = (a+1 -a)/a(a+1) = 1/a(a+1) = LHS.

Thus we proved 1/a -1/(a+1) = 1/a(a+1). Or 1/a(a+1) = 1/a -1/(a+1).

We notice that the denominator of the left side ratio is the least common denominator of 2 irreducible ratios.

We'll suppose that the ratio 1/a(a+1) is the result of addition or subtraction of 2 elementary fractions:

1/a(a+1) = A/a + B/(a+1) (1)

We'll multiply the ratio A/a by (a+1) and we'll multiply the ratio B/(a+1) by a.

1/a(a+1) = [A(a+1) + Ba]/a(a+1)

Since the denominators of both sides are matching, we'll write the numerators, only.

1 = A(a+1) + Ba

We'll remove the brackets:

1 = Aa + A + Ba

We'll factorize by a to the right side:

1 = a(A+B) + A

If the expressions from both sides are equivalent, the correspondent coefficients are equal.

A+B = 0

A = 1

1 + B = 0

B = -1

We'll substitute A and B into the expression (1):

**1/a(a+1) = 1/a - 1/(a+1)**

We remark that we've obtained the request from enunciation.

**The identity has been proved.**