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.

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

Take the case of sum of angles

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

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

Find Tan of both angles

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, $\tan{(90^\circ -C)}$ $=$ $\cot{C}$

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

Expand tan of sum of angles

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

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

Transform cot function in terms of tan function

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.

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

Simplify the trigonometric equation

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.

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

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