How do you prove $tan(3theta)=(3tantheta-tan^3theta)/(1-3tan^2theta)$ ?

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Answer 1 · 2016-10-07T17:22:27

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$tan (3theta) = (3tan theta-tan ^3 theta)/(1-3tan^2 theta)$

Left Side: $=tan (3theta)$

$=tan(2theta+theta)$ ->Use the formulas $tan(A+B)=(tanA+tanB)/(1-tanAtanB)$ and $tan 2A=(2tanA)/(1-tan^2 A)$

$=(tan 2theta+tan theta)/(1-tan 2theta tan theta)$

$=((2tantheta)/(1-tan^2theta)+tan theta)/(1-(2tan theta)/(1-tan^2 theta )tan theta)$

$=((2tan theta+tantheta(1-tan^2theta))/(1-tan^2 theta))/((1-tan^2theta-2tan^2 theta)/(1-tan^2 theta)$

$=(2tan theta+tan theta-tan^3 theta)/(1-tan^2 theta) * (1-tan^2 theta)/(1-tan^2theta-2tan^2 theta)$

$=(3tan theta-tan^3 theta)/cancel (1-tan^2 theta) * (cancel (1-tan^2 theta))/(1-3tan^2 theta)$

$=(3tan theta-tan^3 theta)/(1-3tan^2 theta)$

$:.=$ Right Side

How do you prove tan(3theta)=(3tantheta-tan^3theta)/(1-3tan^2theta)tan(3theta)=(3tantheta-tan^3theta)/(1-3tan^2theta)?