# Law of Tangents

We will discuss here about the law of tangents or the tangent rule which is required for solving the problems on triangle.

In any triangle ABC,

(i) tan ($$\frac{B - C}{2}$$) = ($$\frac{b - c}{b + c}$$) cot $$\frac{A}{2}$$

(ii) tan ($$\frac{C - A}{2}$$)  = ($$\frac{c - a}{c + a}$$) cot $$\frac{B}{2}$$

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

The law of tangents or the tangent rule is also known as Napier’s analogy.

Proof of tangent rule or the law of tangents:

In any triangle ABC we have

⇒ $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$

⇒ $$\frac{b}{c}$$ = $$\frac{sin B}{sin C}$$

⇒ ($$\frac{b - c}{b + c}$$) = $$\frac{sin B - sin C}{sin B + sin C}$$, [Applying Dividendo and Componendo]

⇒ ($$\frac{b - c}{b + c}$$) = $$\frac{2 cos (\frac{B + C}{2}) sin (\frac{B - C}{2})}{2 sin (\frac{B + C}{2}) cos (\frac{B - C}{2})}$$

⇒ ($$\frac{b - c}{b + c}$$) = cot ($$\frac{B + C}{2}$$) tan ($$\frac{B - C}{2}$$)

⇒ ($$\frac{b - c}{b + c}$$) = cot ($$\frac{π}{2}$$ - $$\frac{A}{2}$$) tan ($$\frac{B - C}{2}$$), [Since, A + B + C = π ⇒ $$\frac{B + C}{2}$$ = $$\frac{π}{2}$$ - $$\frac{A}{2}$$]

⇒ ($$\frac{b - c}{b + c}$$) = tan $$\frac{A}{2}$$ tan ($$\frac{B - C}{2}$$)

⇒ ($$\frac{b - c}{b + c}$$) =  $$\frac{tan \frac{B - C}{2}}{cot \frac{A}{2}}$$

Therefore, tan ($$\frac{B - C}{2}$$) = ($$\frac{b - c}{b + c}$$) cot $$\frac{A}{2}$$.                        Proved.

Similarly, we can prove that the formulae (ii) tan ($$\frac{C - A}{2}$$)  = ($$\frac{c - a}{c + a}$$) cot $$\frac{B}{2}$$ and (iii) tan ($$\frac{A - B}{2}$$) = ($$\frac{a - b}{a + b}$$)  cot $$\frac{C}{2}$$.

Alternative Proof law of tangents:

According to the law of sines, in any triangle ABC,

$$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$

Let, $$\frac{a}{sin A}$$ = $$\frac{b}{sin B}$$ = $$\frac{c}{sin C}$$ = k

Therefore,

$$\frac{a}{sin A}$$ = k, $$\frac{b}{sin B}$$ = k and $$\frac{c}{sin C}$$ = k

a = k sin A, b = k sin B and c = k sin C ……………………………… (1)

Proof of formula (i) tan ($$\frac{B - C}{2}$$) = ($$\frac{b - c}{b + c}$$) cot $$\frac{A}{2}$$

R.H.S. = ($$\frac{b - c}{b + c}$$) cot $$\frac{A}{2}$$

= $$\frac{k sin B - k sin C}{k sin B + k sin C }$$ cot $$\frac{A}{2}$$, [Using (1)]

= ($$\frac{sin B - sin C}{sin B + sin C }$$) cot $$\frac{A}{2}$$

= $$\frac{2 sin (\frac{B - C}{2}) cos (\frac{B + c}{2})}{2 sin (\frac{B + C}{2}) cos (\frac{B - c}{2})}$$

= tan ($$\frac{B - C}{2}$$) cot ($$\frac{B + C}{2}$$) cot $$\frac{A}{2}$$

= tan ($$\frac{B - C}{2}$$) cot ($$\frac{π}{2}$$ - $$\frac{A}{2}$$) cot $$\frac{A}{2}$$, [Since, A + B + C = π ⇒ $$\frac{B + C}{2}$$ = $$\frac{π}{2}$$ - $$\frac{A}{2}$$]

= tan ($$\frac{B - C}{2}$$) tan $$\frac{A}{2}$$ cot $$\frac{A}{2}$$

= tan ($$\frac{B - C}{2}$$) = L.H.S.

Similarly, formula (ii) and (iii) can be proved.

Solved problem using the law of tangents:

If in the triangle ABC, C = $$\frac{π}{6}$$, b = √3 and a = 1 find the other angles and the third side.

Solution:

Using the formula, tan ($$\frac{A - B}{2}$$) = ($$\frac{a - b}{a + b}$$) cot $$\frac{C}{2}$$ we get,

tan $$\frac{A - B}{2}$$ = - $$\frac{1 - √3}{1 + √3}$$ cot $$\frac{\frac{π}{6}}{2}$$

tan $$\frac{A - B}{2}$$ = $$\frac{1 - √3}{1 + √3}$$ ∙ cot 15°

tan $$\frac{A - B}{2}$$ = -  $$\frac{1 - √3}{1 + √3}$$ ∙ cot ( 45° - 30°)

tan $$\frac{A - B}{2}$$ = - $$\frac{1 - √3}{1 + √3}$$ ∙ $$\frac{cot 45° cot 30° + 1}{cot 45° - cot 30°}$$

tan $$\frac{A - B}{2}$$ = - $$\frac{1 - √3}{1 + √3}$$ ∙ $$\frac{1 - √3}{1 + √3}$$

tan $$\frac{A - B}{2}$$ = -1

tan $$\frac{A - B}{2}$$ = tan (-45°)

Therefore, $$\frac{A - B}{2}$$ = - 45°

B - A = 90°                            ……………..(1)

Again, A + B + C = 180°

Therefore, A + 8 = 180° - 30° = 150° ………………(2)

Now, adding (1) and (2) we get, 2B = 240°

B = 120°

Therefore, A = 150° - 120° = 30°

Again, $$\frac{a}{sin A}$$ = $$\frac{c}{sin C}$$

Therefore, $$\frac{1}{sin 30°}$$ = $$\frac{c}{sin 30°}$$

c = 1

Therefore, the other angles of the triangle are 120° or, $$\frac{2π}{3}$$; 30° or, $$\frac{π}{6}$$; and the length of the third side = c = 1 unit.

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