How do you simplity #6cot4theta-tan3theta# to trigonometric functions of #theta#?

1 Answer
Mar 17, 2018

#((36tan^2theta-6tan^4theta-6)/(4tantheta-4tan^3theta))-((3tantheta-tan^3theta)/(1-3tan^2theta))#

Explanation:

Please have a look at double angle tan and cot identities and cot and tan sum identities before you begin, rest is simplification

Start:
#6cot(2theta+2theta)-tan(2theta+theta)#
#6((1-(1-tan^2theta)/(2tantheta)*(1-tan^2theta)/(2tantheta))/((1-tan^2theta)/(2tantheta)+(1-tan^2theta)/(2tantheta)))-(((2tantheta)/(1-tan^2theta)+tantheta)/(1-(2tan^2theta)/(1-tan^2theta)))#

#6((1-(tan^4theta-2tan^2theta+1)/(4tan^2theta))/((2-2tan^2theta)/(2tantheta)))-(((2tantheta)/(1-tan^2theta)+tantheta)/((1-tan^2theta)/(1-tan^2theta)-(2tan^2theta)/(1-tan^2theta)))#

#6(((4tan^2theta)/(4tan^2theta)-(tan^4theta-2tan^2theta+1)/(4tan^2theta))/((2-2tan^2theta)/(2tantheta)))-(((2tantheta)/(1-tan^2theta)+tantheta)/((1-3tan^2theta)/(1-tan^2theta)))#

#6(((6tan^2theta-tan^4theta-1)/(4tan^2theta))/((2-2tan^2theta)/(2tantheta)))-(((2tantheta)/(1-tan^2theta)+(tantheta-tan^3theta)/(1-tan^2theta))/((1-3tan^2theta)/(1-tan^2theta)))#

#6(((6tan^2theta-tan^4theta-1)/(4tan^2theta))/((2-2tan^2theta)/(2tantheta)))-(((3tantheta-tan^3theta)/(1-tan^2theta))/((1-3tan^2theta)/(1-tan^2theta)))#

#6((6tan^2theta-tan^4theta-1)/(cancel(2tantheta)*2tantheta)*cancel(2tantheta)/(2-2tan^2theta))-(((3tantheta-tan^3theta)/(1-tan^2theta))/((1-3tan^2theta)/(1-tan^2theta)))#

#6((6tan^2theta-tan^4theta-1)/(4tantheta-4tan^3theta))-(((3tantheta-tan^3theta)/(1-tan^2theta))/((1-3tan^2theta)/(1-tan^2theta)))#

#6((6tan^2theta-tan^4theta-1)/(4tantheta-4tan^3theta))-(((3tantheta-tan^3theta)/cancel(1-tan^2theta))*((cancel(1-tan^2theta))/(1-3tan^2theta)))#

#((36tan^2theta-6tan^4theta-6)/(4tantheta-4tan^3theta))-((3tantheta-tan^3theta)/(1-3tan^2theta))#