$A$, $B$ 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.
$\,\,\, \therefore \,\,\,\,\,\,$ $\sum \tan{A}\tan{B}$ $\,=\,$ $1$
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved