Tangent of sum of Two angles

tan of sum 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

Construction of triangle to derive expansion of tangent of compound angle that formed by the addition of two angles

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.

Construction of triangle to derive expansion of tangent of compound angle that formed by the addition of two angles

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

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 $NMP$.

Construction of triangle to derive expansion of tangent of compound angle that formed by the addition of two angles

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

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

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 $\Delta NMP$

Construction of triangle to derive expansion of tangent of compound angle that formed by the addition of two angles

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 $\Delta PMQ$

Construction of triangle to prove cosine of compound angle that formed by the addition of two angles

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.

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

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