Expert: Josh - 3/28/2006

Question
I have a worksheet that asks me to Prove each Identity of:

(tan a - tan b)/(tan a + tan b) =
sin (a-b) / sin (a + b)

Not sure where to go with this?

Answer
Hi Eric,

This question requires you knowing the following entities:
sin(a-b)=sin(a)cos(b)-cos(a)sin(b) ...[#1]
sin(a+b)=sin(a)cos(b)+cos(a)sin(b) ...[#2]

To answer this question,
Rewrite the right hand side (RHS) of the equation, substituting [#1],[#2] for sin(a-b) and sin(a+b).

We get sin(a-b)/sin(a+b)=[sin(a)cos(b)-cos(a)sin(b)]/[sin(a)cos(b)+cos(a)sin(b)] ...[#3]

To show that this is equivalent to the left hand side (LHS) of the original expression, we need to introduce the "tangent" terms.

To do this, divide the numerator and denominator both by cos(a)cos(b). Remember that sin(a)/cos(a)=tan(a).

So, picking up from line [#3], in the numerator we have
sin(a)cos(b)/[cos(a)cos(b)]=tan(a), since the "cos(b)" term cancels out. The next term in the numerator cos(a)sin(b)/[cos(a)cos(b)]=tan(b).

Do the same for the denominator, which contains the expression [sin(a)cos(b)+cos(a)sin(b)] and you'll get [tan(a)+tan(b)].