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tanθ + tan2θ + √3 tanθ tan2θ = √3 Find tanθ ? Please help.

tanθ + tan2θ + √3 tanθ tan2θ = √3  Find tanθ ? Please help.

Grade:11

2 Answers

Sreevignesh
22 Points
7 years ago
I can't insert theta so.... I am using x instead of theta
Tan x+tan 2x+ √3 tan x tan 2x =√3
                            Tan x + tan 2x = √ 3- √3 tan x tan 2 x.    
                            Tan x +tan 2 x= √3(1-tan x tan 2x)
                                             √3.  = (tan x+ tan 2x)/(1-tan x tan 2 x)
                                      √3 = tan ( x+2x).           [Tan (x+2x)=(tan x+ tan 2x)/(1-tan x tan 2x)]
                                      √3 = tan 3 x
                                     3x.  = tan ^-1 ( √3)
                                     3x.   =  π/3
                                       x.   = π/9 
                            Now,  tan x = tan π/9 
                                                = tan 20° = 0.3639
 
      
Kyra
25 Points
7 years ago
Take root 3tantheta tan 2theta to the right hand side and then take root 3 common and then bring 1+tan theta tan2theta in denominator then tan3theta is equal to root 3.

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