Question

For A = 60° and B = 30°, verify that tan (A – B) = `(tan A - tan B)/(1 + tan A tan B)`.

Answer 1

1. Evaluate the Left-Hand Side (LHS)

Substitute the values of A and B into the expression tan (A – B):

LHS = tan (60° – 30°)

LHS = tan 30°

From trigonometric tables, the value of tan 30° is:

LHS = `1/sqrt(3)`

2. Evaluate the Right-Hand Side (RHS)

Substitute the values of A and B into the expression `(tan A - tan B)/(1 + tan A tan B)`

RHS = `(tan 60^circ -  tan 30^circ)/(1 + tan 60^circ tan 30^circ)`

Recal that `tan 60^circ = sqrt(3)` and `tan 30^circ = 1/sqrt(3)`.

RHS = `(sqrt(3) - 1/sqrt(3))/(1 + (sqrt(3) xx 1/sqrt(3))`

3. Simplify the RHS

Simplify the numerator and the denominator separately:

Numerator: `sqrt(3) - 1/sqrt(3)`

= `((sqrt(3) xx sqrt(3)) - 1)/sqrt(3)`

= `(3 - 1)/sqrt(3)`

= `2/sqrt(3)`

Denominator: `1 + (sqrt(3) xx 1/sqrt(3))`

= 1 + 1

= 2

Now combine them:

RHS = `(2/sqrt(3))/2`

= `2/sqrt(3) xx 1/2`

= `1/sqrt(3)`

Since the LHS = RHS = `1/sqrt(3)`, the identity is verified for the given angles.