How do you simplify #(4sinxcosx-2sinxsecx)/(2tanx)#?

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Answer 1 · 2016-09-15T09:49:04

#(4sinxcosx-2sinxsecx)/(2tanx)=cos2x#

#(4sinxcosx-2sinxsecx)/(2tanx)#

= #(4sinxcosx-2sinx/cosx)/(2sinx/cosx)#

= #(4sinxcosx)/(2sinx/cosx)-(2sinx/cosx)/(2sinx/cosx)#

= #(4sinxcos^2x)/(2sinx)-1#

= #2cos^2x-1#

= #cos2x#

Answer 2 · 2016-09-15T11:49:35

#LHS=(4sinxcosx-2sinxsecx)/(2tanx)#

#=(2sinx/cosx(2cos^2x-1))/(2tanx)#

#=(tanx(2cos^2x-1))/(tanx)#

#=(tanx(cos2x))/(tanx)=cos2x#