Law Of Tangents: Definition, Proof, Formula and Examples

In trigonometry, the law of tangents or tangent rule is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides. The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known. In real life, we use the Law of Tangents to calculate the angle of celestial objects such as stars, moon, etc.

In this article, we will cover the concept of the Tangent rule. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of five questions have been asked on this concept

Tangent Rule: Definition

The tangent rule states that the ratio of the sum and difference of any two angles of a triangle is equal to the ratio of tangents of half the sum and the difference of angle opposites to the side. The tangent rule describes the relationships between the tangent of two angles of the triangle and the length of the opposite side.

Tangent Rule Formula

The law of tangents or tangent rule is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides. The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known. The law of tangents can be used to compute the angles of a triangle in which two sides a and b and the enclosed angle C are given.

For any ΔABC,

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

Derivation of Tangent rule formula

To prove this, we will be using the sine rule and formula of sum/difference into the product of sine and cosine.

From the sine rule, we have

$
\begin{aligned}
& \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\
& \Rightarrow \frac{\sin B}{\sin C}=\frac{b}{c}
\end{aligned}
$
Using Componendo and dividendo theorem

$
\Rightarrow \frac{\sin B-\sin C}{\sin B+\sin C}=\frac{b-c}{b+c}
$

formula of sum/Difference into products of sine

$
\begin{aligned}
& \Rightarrow \quad \frac{2 \cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \sin \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{2 \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \\
& \Rightarrow \quad \cot \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \\
& \Rightarrow \quad \tan \frac{\mathrm{A}}{2} \tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \\
& {\left[\because \frac{\mathrm{B}+\mathrm{C}}{2}=\frac{\pi-\mathrm{A}}{2} \Rightarrow \cot \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right)=\tan \frac{\mathrm{A}}{2}\right]} \\
& \Rightarrow \quad \frac{\tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{\cot \frac{\mathrm{A}}{2}}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \\
& \Rightarrow \quad \tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \cot \frac{\mathrm{A}}{2}
\end{aligned}
$

By using the same method, other formulas can also be proved

So, In triangle ABC, we have

$\begin{aligned} & \tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b} \cot \frac{C}{2} \\ & \tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2} \\ & \tan \left(\frac{C-A}{2}\right)=\frac{c-a}{c+a} \cot \frac{B}{2}\end{aligned}$

These formulas are also known as the Tangent rule. This is useful in calculating the remaining parts of the triangle when two sides and the included angle are given.

The law of tangent( tan rule ) can be applied in the following cases:

1) When two sides and one angle is given

2) Two angles and one side is given

3) Three sides are given

4) Two sides and the angle between them are given

Summary

The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known. The law of tangent is not only used to derive various formulas in trigonometry but also helps in the fields of physics and astrology. Knowledge of tangent rules is useful in solving complex problems.

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Solved Examples Based on Tangent Rule

Example 1: In a $\triangle A B C$, the sides $\mathrm{a}=4, \mathrm{~b}=2$ and $\angle c=90^{\circ}$. To find the value of $A-B$.
Solution
Given that,

$
a=4, b=2 \text { and } \angle c=90^{\circ}
$
We know that,

$
\begin{aligned}
& \angle A+\angle B+\angle C=180^{\circ} \\
& \angle A+\angle B=180^{\circ}-\angle C \\
& \angle A+\angle B=90^{\circ}
\end{aligned}
$
By the law of tangents,

$
\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}
$
Putting the value in the equation,

$
\begin{aligned}
& \frac{4-2}{4+2}=\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}\left(90^{\circ}\right)} \\
& \frac{1}{3}=\frac{\tan \frac{1}{2}(A-B)}{\tan \left(45^{\circ}\right)} \\
& \frac{1}{3} \tan \left(45^{\circ}\right)=\tan \frac{1}{2}(A-B)
\end{aligned}
$
Solving,

Since x is in the first quadrant, $\cos x$ is positive. Thus,

Substituting the value of $\sec x=\frac{4}{3}$ to get,
$\tan x= \pm \sqrt{\left(\frac{4}{3}\right)^2-1}$ $\tan x=\sqrt{\frac{7}{9}}$

Hence, the answer is $\sqrt{\frac{7}{9}}$

Example 2: With the usual notation, in $\triangle A B C$, if $\angle A+\angle B=120^{\circ}, a=\sqrt{3}+1$ and $b=\sqrt{3}-1$, then the ratio $\angle A: \angle B$, is:
Solution: We know the Addition Formulae $\sin (A+B)=\sin A \cos B+\cos A \sin B$
Now,
Given $\angle A+\angle B=120^{\circ}$
From the concept

Using Napiers Analogy

$
\begin{aligned}
\tan \left(\frac{A-B}{2}\right) & =\frac{a-b}{a+b} \cot \frac{C}{2} \\
& =\frac{\sqrt{3}+1-\sqrt{3}+1}{2 \sqrt{3}} \cot 30^{\circ} \\
& =1
\end{aligned}
$

we get,

$
\because \tan 45^{\circ}=1
$
So,

$
\begin{aligned}
& \frac{A-B}{2}=45 \\
& \Rightarrow A-B=90
\end{aligned}
$
From (1) and (2)

$
\Rightarrow A=105^{\circ}, B=15^{\circ}
$
Hence, the answer is $7: 1$

Example 3: If $
\cot \frac{B+C}{2} \cdot \tan \frac{B-C}{2}=x, \text { then } \mathrm{x} \text { equals. }
$

Solution: We know the tangent rule

$
\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}
$
Multiply both sides by tan(A/2) we get,

$
\tan \frac{A}{2} \tan \frac{B-C}{2}=\frac{b-c}{b+r}
$
From triangle ABC ,

$
\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& \mathrm{A}=\pi-(\mathrm{B}+\mathrm{C})
\end{aligned}
$
Putting the value of $A$ in the equation,

$
\tan \frac{\pi-(B+C)}{2} \tan \frac{B-C}{2}=\frac{b-c}{b+c}
$
We know $\tan (\pi / 2-\mathrm{A})=\cot \mathrm{A}$,

$
\cot \frac{B+C}{2} \tan \frac{B-C}{2}=\frac{b-c}{b+c}
$
Hence, $
x=\frac{b-c}{b+c}
$

Hence, the answer is $\frac{b-c}{b+c}$

Example 4: Solve: $\cot \frac{A+B}{2} \cdot \tan \frac{A-B}{2}=$ ?

Solution: According to Napier's Analogy, for any $\triangle A B C$

$
\begin{aligned}
& \tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b} \cot \frac{C}{2} \\
& \tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2} \\
& \tan \left(\frac{C-A}{2}\right)=\frac{c-a}{c+a} \cot \frac{B}{2}
\end{aligned}
$
We know $\tan \frac{A-B}{2}=\frac{a-b}{a+b} \cot \frac{C}{2} \ldots(i)$

$
\begin{aligned}
& A+B+C=\pi \\
& \frac{1}{2}(A+B)=\frac{1}{2}(\pi-C) \\
& \tan \frac{(A+B)}{2}=\tan \frac{(\pi-C)}{2} \\
& \tan \frac{(A+B)}{2}=\cot \frac{(C)}{2}
\end{aligned}
$

by equation (i)

$
\begin{aligned}
& \tan \frac{A-B}{2}=\frac{a-b}{a+b} \tan \frac{(A+B)}{2} \\
& \cot \frac{A+B}{2} \cdot \tan \frac{A-B}{2}=\frac{a-b}{a+b}
\end{aligned}
$
Hence, the answer is $\frac{a-b}{a+b}$