Trigonometry/Law of Tangents

For any triangle with angles $A,B,C$ and corresponding opposite side lengths $a,b,c$ , the Law of Tangents states that

$\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}$

Corresponding identities also hold in terms of $b,c,B,C$ and in terms of $c,a,C,A$ .

When to use it

This formula is nothing like as important as the Law of Sines or the Law of Cosines, which is why we have put it and its proof in the reference section. This formula may be found in your formula book. We're including it and the proof of it 'for completeness'.

Its main use, as far as we are concerned, is that to prove it is a good piece of practice in algebra with trig.

Proof

From the sine theorem,

$\frac{\sin (A)}{a}=\frac{\sin (B)}{b}$

Hence

$\frac{a-b}{a+b}=\frac{\sin (A)-\sin (B)}{\sin (A)+\sin (B)}=\frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}$

We're going to get you to spell out the details of each step in the above:

Template:ExerciseRobox Let's introduce a constant $k$

$k=\frac{a}{\sin (A)}=\frac{b}{\sin (B)}$

Now express $a,b$ in terms of $k,\sin (A),\sin (B)$ . Use these expressions for $a,b$ to remove $a,b$ entirely from $\frac{a-b}{a+b}$ .

Finish by cancelling the $k$'s.

Template:ExerciseRobox Use the sum to products formula for the second step, or express $A,B$ as

$A=\frac{A+B}{2}+\frac{A-B}{2}$

and

$B=\frac{A+B}{2}-\frac{A-B}{2}$

and use sine addition formula to transform the numerator and denominator into products. It might look a bit scary at first, but do it right and you'll see that lots of terms cancel out. This step, using the sine addition formula, is actually practice in the derivation of the sum to products formula.

In the penultimate expression (i.e. the last but one expression), look for terms like:

$\frac{\sin (x)}{\cos (x)}$

with $x$ the same expression. These can be replaced by $\tan (x)$ , where $x$ is that expression.

Remember that if $\frac{a}{b}=\tan (\theta )$ then it follows that $\frac{b}{a}=\frac{1}{\tan (\theta )}$ .