# Tangent of sum of Two angles

Tangent of a compound angle that formed by the summation of two angles of a right angled triangle can be expressed in terms of same trigonometric ratio but with same two angles.

For example, $A$ and $B$ are two angles and summation of them $(A+B)$ is a compound angle. It is also an angle of the right angled triangle. $\tan(A+B)$ is tangent of compound angle and it is expressed in terms of tan with $A$ and $B$ angles.

## Formula

$$\tan (A+B) \,\,=\,\, \frac{\tan A \,+\, \tan B}{1 \,-\, \tan A . \tan B}$$

### Proof

To understand the derivation of tan of summation of two angles, a right angled triangle is constructed by a compound angle which is formed by the addition of two angles.

Firstly, understand our step by step construction process of a triangle to study its properties and they are used in deriving the expression of tangent of compound angle in mathematical form.

1

#### According to $\Delta SMQ$

The compound angle $A+B$ is the angle of the right angled triangle $SMQ$. Express tangent of compound angle in its mathematical form.

$$\tan(A+B) \,=\, \frac{QS}{MS}$$

The length of the side $\overline{QS}$ can be expressed as the addition of the sides $\overline{QU}$ and $\overline{US}$. Similarly, the length of the side $\overline{MS}$ can also be expressed as the subtraction of length of the side $\overline{SN}$ from side $\overline{MN}$.

$$\implies \tan(A+B) \,=\, \frac{QU \,+\, US}{MN \,-\, SN}$$

The length of the side $\overline{US}$ is exactly equal to the length of the side $\overline{PN}$.

$$\implies \tan(A+B) \,=\, \frac{QU \,+\, PN}{MN \,-\, SN}$$

The length of the side $\overline{SN}$ is equal to length of the side $\overline{UP}$. So, replace $SN$ by $UP$ in denominator of the $\tan(A+B)$

$$\implies \tan(A+B) \,=\, \frac{QU \,+\, PN}{MN \,-\, UP}$$

2

#### According to $\Delta NMP$

Express $\tan$ of angle $A$ in terms of ratio of two sides by considering right angled triangle $NMP$.

$$\tan A = \frac{PN}{MN}$$

$$\implies PN = MN \times \tan A$$

Replace the value of length of side $\overline{PN}$ in $\tan(A+B)$ expression.

$$\implies \tan(A+B) \,=\, \frac{QU \,+\, MN \times \tan A}{MN \,-\, UP}$$

Take length of the side $\overline{MN}$ common in both numerator and denominator.

$$\implies \tan(A+B) \,=\, \frac{MN \times \Bigg( \dfrac{QU}{MN} \,+\, \tan A \Bigg) }{MN \times \Bigg( 1 \,-\, \dfrac{UP}{MN} \Bigg)}$$

$$\require{cancel} \implies \tan(A+B) \,=\, \frac{\cancel{MN}}{\cancel{MN}} \times \frac{\Bigg(\dfrac{QU}{MN} \,+\, \tan A \Bigg)}{\Bigg( 1 \,-\, \dfrac{UP}{MN} \Bigg)}$$

$$\implies \tan(A+B) \,=\, \frac{\Bigg(\dfrac{QU}{MN} \,+\, \tan A \Bigg)}{\Bigg( 1 \,-\, \dfrac{UP}{MN} \Bigg)}$$

3

#### According to $\Delta PQU$ and $\Delta NMP$

Express $\tan A$ in terms of ratio of two sides of the right angled triangle $PQU$.

$$\tan A = \frac{UP}{QU}$$

$$\implies UP = QU \times \tan A$$

Replace length of the side $\overline{UP}$ by its value in $\tan(A+B)$ expression.

$$\implies \tan(A+B) \,=\, \frac{\Bigg( \dfrac{QU}{MN} \,+\, \tan A \Bigg) }{\Bigg( 1 \,-\, \dfrac{QU \times \tan A}{MN} \Bigg)}$$

$$\implies \tan(A+B) \,=\, \frac{\Bigg( \dfrac{QU}{MN} \,+\, \tan A \Bigg) }{\Bigg( 1 \,-\, \dfrac{QU}{MN} \times \tan A \Bigg)}$$

Express $\cos A$ in terms of sides of the right angled triangle $PQU$.

$$\cos A = \frac{QU}{PQ}$$

Express $\cos A$ in terms of sides of the right angled triangle $NMP$.

$$\cos A = \frac{MN}{MP}$$

The two ratios are values of $\cos A$. So, they both are equal in value.

$$\frac{QU}{PQ} = \frac{MN}{MP}$$

$$\implies \frac{QU}{MN} = \frac{PQ}{MP}$$

Now replace ratio of $QU$ to $MN$ by the ratio of $PQ$ to $MP$ in expression of $\tan(A+B)$

$$\implies \tan(A+B) \,=\, \frac{\Bigg( \dfrac{PQ}{MP} \,+\, \tan A \Bigg) }{\Bigg( 1 \,-\, \dfrac{PQ}{MP} \times \tan A \Bigg)}$$

4

#### According to $\Delta PMQ$

Express $\tan$ of angle $B$ in terms of sides of the right angled triangle $PMQ$.

$$\tan B = \frac{PQ}{MP}$$

Replace the ratio of $PQ$ to $MP$ by $\tan B$ in the expression of $\tan(A+B)$ to get required expansion.

$$\implies \tan(A+B) \,=\, \frac{\tan B \,+\, \tan A}{1 \,-\, \tan B \times \tan A}$$

It can be written as follows.

$$\therefore \,\, \tan(A+B) \,=\, \frac{\tan A \,+\, \tan B}{1 \,-\, \tan A . \tan B}$$

##### Alternative forms

The two angles which form a compound angle by summation can be denoted by any two symbols but the expansion of the tangent of summation of two angles is same.

If $x$ and $y$ are two angles, the trigonometric identity is written as follows.

$$\tan (x+y) \,\,=\,\, \frac{\tan x \,+\, \tan y}{1 \,-\, \tan x . \tan y}$$

If $\alpha$ and $\beta$ are two angles, the angle sum trigonometric identity is expressed as follows.

$$\tan (\alpha + \beta) \,\,=\,\, \frac{\tan \alpha \,+\, \tan \beta}{1 \,-\, \tan \alpha . \tan \beta}$$