Math Doubts

Tan of angle sum identity

Expansion form

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

Simplified form

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

Introduction

The tan of angle sum identity is called as tan of sum of two angles identity or tan of compound angle identity. It is mainly used in mathematics in two cases possibly.

  1. To expand tan of sum of two angles as the quotient of sum of tangents of angles by the subtraction of products of tangents of both angles from one.
  2. To simplify the quotient of sum of tangents of angles by the subtraction of products of tangents of both angles from one as tan of sum of two angles.

Formula

The tan of angle sum identity is written in several ways in which $\tan{(A+B)}$, $\tan{(x+y)}$ and $\tan{(\alpha+\beta)}$ are popular in the world. You can write it in terms of any two angles.

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

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

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

Proof

You have learned the tan of sum of two angles formula and let’s learn how the tan of angle sum identity is derived in mathematical form by the geometrical approach.



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