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

Question

50072 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-07T17:22:27

see below

#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