$(1).\,\,$ $\tan{(a+b)}$ $\,=\,$ $\dfrac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$

$(2).\,\,$ $\tan{(x+y)}$ $\,=\,$ $\dfrac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

Let $a$ and $b$ be two variables. Let’s assume that they represent two angles. The sum of two angles is written as $a+b$ in mathematics. It is actually a compound angle. The tangent of a compound angle $a$ plus $b$ is expressed as $\tan{(a+b)}$ mathematically.

The tan of the sum of angles $a$ and $b$ is equal to the quotient of the sum of the tangents of angles $a$ and $b$ by the subtraction of the product of tangents of angles $a$ and $b$ from one.

$\tan{(a+b)}$ $\,=\,$ $\dfrac{\tan{a}+\tan{b}}{1-\tan{a} \times \tan{b}}$

The above mathematical equation is called the tangent of angle sum trigonometric identity in mathematics.

The tan angle sum trigonometric identity is used possibly in two cases in mathematics.

The tan of the sum of two angles is expanded as the quotient of the sum of the tangents of angles by the subtraction of the product of tangents of angles from one.

$\implies$ $\tan{(a+b)}$ $\,=\,$ $\dfrac{\tan{(a)}+\tan{(b)}}{1-\tan{(a)}\tan{(b)}}$

The quotient of the sum of the tangents of angles by the subtraction of the product of tangents of angles from one is simplified as the tan of the sum of two angles.

$\implies$ $\dfrac{\tan{(a)}+\tan{(b)}}{1-\tan{(a)}\tan{(b)}}$ $\,=\,$ $\tan{(a+b)}$

The angle sum trigonometric rule in tangent function is written in several forms but it is expressed popularly in the following three forms.

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

$(2).\,\,$ $\tan{(x+y)}$ $\,=\,$ $\dfrac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

$(3).\,\,$ $\tan{(\alpha+\beta)}$ $\,=\,$ $\dfrac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$

Learn how to derive the tan of angle sum trigonometric identity by a geometric method in trigonometry.

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