$A$, $B$ 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.

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

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Oct 22, 2024

Oct 17, 2024

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved