If $A+B+C = 90^\circ$, then find $\sum \tan{A}\tan{B}$

$A$, $B$ and $C$ are three angles and it is given in this problem that the sum of them is equal to $90^\circ$.

$A+B+C = 90^\circ$

The value of $\sum \tan{A}\tan{B}$ has to find in this case.

The mathematical meaning of $\sum \tan{A}\tan{B}$ is the sum of product of two tan functions.

$\sum \tan{A}\tan{B}$ $=$ $\tan{A}\tan{B}$ $+$ $\tan{B}\tan{C}$ $+$ $\tan{C}\tan{A}$

$A+B+C = 90^\circ$

$\implies$ $A+B = 90^\circ -C$

Take tan both sides to find the value of tan of both angles.

$\implies$ $\tan{(A+B)} = \tan{(90^\circ -C)}$

According to allied angle identities, $\tan{(90^\circ -C)}$ $=$ $\cot{C}$

$\implies$ $\tan{(A+B)} = \cot{C}$

As per angle sum identities, expand the tan of sum of two angles.

$\implies$ $\dfrac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$ $\,=\,$ $\cot{C}$

Tangent and Cotangent functions are reciprocal functions. So, convert cot function in terms of tan function as per reciprocal relation of cot function with tan function.

$\implies$ $\dfrac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$ $\,=\,$ $\dfrac{1}{\tan{C}}$

Use cross multiplication technique to simplify the trigonometric equation.

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

$\implies$ $\tan{A}\tan{C}$ $+$ $\tan{B}\tan{C}$ $\,=\,$ $1-\tan{A}\tan{B}$

$\implies$ $\tan{A}\tan{B}$ $+$ $\tan{A}\tan{C}$ $+$ $\tan{B}\tan{C}$ $\,=\,$ $1$

$\implies$ $\tan{A}\tan{B}$ $+$ $\tan{B}\tan{C}$ $+$ $\tan{A}\tan{C}$ $\,=\,$ $1$

$\implies$ $\tan{A}\tan{B}$ $+$ $\tan{B}\tan{C}$ $+$ $\tan{C}\tan{A}$ $\,=\,$ $1$

The trigonometric expression can be written in short form.

$\,\,\, \therefore \,\,\,\,\,\,$ $\sum \tan{A}\tan{B}$ $\,=\,$ $1$