Law of tangents is a less known formula when compared with Laws of sines and cosines. Law of tangents establishes a ratio relationship between the sum and difference of two sides of a triangle to the tangents of half the sum and differences of the angles opposite to the sides. It can also be used to solve triangles when two sides and the included angle are given. Thus in SAS situations, Law of tangents can be used instead of Law of cosines. Like laws of sines and cosines law of tangents is also used in proving identities related to the measures of a triangle.
The law of tangents for the ΔABC consists of three formulas as follows:

1. $\frac{a+b}{a-b}$ = $\frac{tan\frac{A+B}{2}}{tan\frac{A-B}{2}}$

2. $\frac{b+c}{b-c}$ = $\frac{tan\frac{B+C}{2}}{tan\frac{B-C}{2}}$

3. $\frac{c+a}{c-a}$ = $\frac{tan\frac{C+A}{2}}{tan\frac{C-A}{2}}$

Where a, b and c are the three sides of a triangle. And A, B and C are angle opposite to these sides.

Let us look into the proof of law of tangents and solve few problems using this property.

The law of tangents is proved using sine rule and sine sum and difference rules.
The Law of sines states that in any triangle the measure of a side bears a constant ratio to the sine of the angle opposite to it. Thus we have,

$\frac{a}{sinA}$ = $\frac{b}{sinB}$ = $\frac{c}{sinC}$ = k

Thus we have a = k.sin A, b = k.sin B and c = k.sin C.
The Law of tangents states

$\frac{a+b}{a-b}$ = $\frac{tan\frac{A+B}{2}}{tan\frac{A-B}{2}}$
Considering the left side of identity,

$\frac{a+b}{a-b}$ = $\frac{ksinA+ ksinB}{ksinA-ksinC}$ Applying law of sines

= $\frac{k(sinA+sinB)}{k(sinA-sinB)}$ Common factor taken out

= $\frac{sinA+sinB}{sinA-sinB}$

= $\frac{2sin(\frac{A+B}{2})cos(\frac{A-B}{2})}{2cos(\frac{A+B}{2})sin(\frac{A-B}{2})}$ Sum and difference laws for sine function.

= $\frac{tan(\frac{A+B}{2})}{tan(\frac{A-B}{2})}$ Simplified using Quotient identities

= Right side of the identity.

The other two versions can also be proved considering the corresponding ratios and using the related sum and difference formulas.

It is sufficient to remember one of the formulas. The other two can be suitably got by replacing the sides and angles in a symmetrical order.
Law of tangents can be used to solve triangles in SAS situation, that is when two sides and the included angle are known.

Solve ΔABC for the unknown measures.

Law of Tangents Problems

In ΔABC, the measure of ∠C is given as 70º.
A + B = 180 - C = 180 - 70 = 110º.

$\frac{a+b}{a-b}$ = $\frac{tan\frac{A+B}{2}}{tan\frac{A-B}{2}}$                                                  Law of Tangents

$\frac{9+7}{9-7}$ = $\frac{tan\frac{110}{2}}{tan\frac{A-B}{2}}$

8 = $\frac{tan55}{tan\frac{A-B}{2}}$

tan $\frac{A-B}{2}$ = $\frac{tan55}{8}$                                               Simplified and

solved for tan $\frac{A-B}{2}$

                               ≈ 0.1785

      $\frac{A-B}{2}$ = tan-1 0.1785 = 10.1º                          Evaluated using calculator and angle measure rounded to the tenth.

Now the angles A and B can be found solving the simultaneous equations,
A + B = 110
A - B = 20.2º  
A = 65.12º and B = 44.88º

The measure of side c can be calculate using sin rule as follows:

$\frac{a}{sinA}$ = $\frac{c}{sinC}$

c = Sin C x $\frac{a}{sinA}$

   = Sin 70 x $\frac{9}{sin 65.12}$

   ≈ 9.322 cm

Now let us prove an identity using Law of tangents.

Verify the identity cot $\frac{C}{2}$ = $\frac{a+b}{a-b}$ . tan $\frac{A-B}{2}$

Consider the law of tangents,

$\frac{tan\frac{A+B}{2}}{tan\frac{A-B}{2}}$ = $\frac{a+b}{a-b}$

Multiplying the equation by tan $\frac{A-B}{2}$  (The difference between two angles in a triangle A - B ≠ 0).

tan $\frac{A+B}{2}$ = $\frac{a+b}{a-b}$ . tan $\frac{A-B}{2}$

tan ($\frac{180-C}{2}$) = $\frac{a+b}{a-b}$ . tan $\frac{A-B}{2}$              A + B + C = 180º.

tan (90 - $\frac{C}{2}$) = $\frac{a+b}{a-b}$ . tan $\frac{A-B}{2}$ 

cot $\frac{C}{2}$ = $\frac{a+b}{a-b}$ . tan $\frac{A-B}{2}$                       cofunction identity tan (90 - θ) = cot θ.