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

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

The cot of angle sum identity is called as cot of sum of two angles identity or cot of compound angle identity. It is usually used in two cases in mathematics.

- To expand cot of sum of two angles as the quotient of subtraction of one from products of cotangents of angles by the sum of cotangents of the angles.
- To simplify the subtraction of one from products of cotangents of angles by the sum of cotangents of the angles as the cot of sum of two angles.

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

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

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

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

You learned how to expand cot of sum of two angles function and also learned how to write its expansion in simplified form. It is your time to learn how to derive cot of angle sum identity in trigonometric mathematics.

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