# Tan X Tan ( X + π 3 ) + Tan X Tan ( π 3 − X ) + Tan ( X + π 3 ) Tan ( X − π 3 ) = − 3 - Mathematics

Numerical

$\tan x \tan\left( x + \frac{\pi}{3} \right) + \tan x \tan\left( \frac{\pi}{3} - x \right) + \tan\left( x + \frac{\pi}{3} \right)\tan\left( x - \frac{\pi}{3} \right) = - 3$

#### Solution

$\frac{\pi}{3} = 60°$
$LHS = \text{ tan } x \tan\left( x + 60° \right) + \text{ tan } x \tan\left( x - 60° \right) + \tan\left( x + 60° \right)\tan\left( x - 60° \right)$
$= \text{ tan } x\left( \frac{\text{ tan } x + \sqrt{3}}{1 - \sqrt{3}\text{ tan } x} \right) + \text{ tan } x\left( \frac{\text{ tan } x - \sqrt{3}}{1 + \sqrt{3}\text{ tan } x} \right) + \left( \frac{\text { tan } x + \sqrt{3}}{1 - \sqrt{3}\text{ tan } x} \right)\left( \frac{\text{ tan } x - \sqrt{3}}{1 + \sqrt{3}\text{ tan } x} \right)$
$\left[ \tan\left( A + B \right) = \frac{\text{ tan } A + \text{ tan } B}{1 - \text{ tan }A\text{ tan } B} \text{ and } \tan\left( A - B \right) = \frac{\text{ tan } A - \text{ tan } B}{1 + \text{ tan } A\text{ tan } B} \right]$
$= \frac{\left( 1 + \sqrt{3}\text{ tan } x \right)\text { tan } x\left( \text { tan } x + \sqrt{3} \right) + \left( 1 - \sqrt{3}\text { tan } x \right)\text { tan } x\left( \text { tan } x - \sqrt{3} \right) + \tan^2 x - 3}{\left( 1 + \sqrt{3}\text{ tan } x \right)\left( 1 - \sqrt{3}\text{ tan } x \right)}$
$= \frac{\left( \text{ tan } x + \sqrt{3} \tan^2 x \right)\left( \text{ tan } x + \sqrt{3} \right) + \left( \text{ tan } x - \sqrt{3}\text { tan } x \right)\left( \text{ tan } x - \sqrt{3} \right) + \tan^2 x - 3}{1 - 3 \tan^2 x}$

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.2 | Q 4 | Page 36