Solve #1/(tan2x-tanx)-1/(cot2x-cotx)=1#?

Question

2065 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-10T14:50:44

#1/(tan2x-tanx)-1/(cot2x-cotx)=1#

#=>1/(tan2x-tanx)-1/(1/(tan2x)-1/tanx)=1#

#=>1/(tan2x-tanx)+1/(1/(tanx)-1/(tan2x))=1#

#=>1/(tan2x-tanx)+(tanxtan2x)/(tan2x-tanx)=1#

#=>(1+tanxtan2x)/(tan2x-tanx)=1#

#=>1/tan(2x-x)=1#

#=>tan(x)=1=tan(pi/4)#

#=>x=npi+pi/4#

Answer 2 · 2017-11-10T14:58:37

#x=npi+pi/4#

#tan2x-tanx=(sin2x)/(cos2x)-sinx/cosx=(sin2xcosx-cos2xsinx)/(cos2xcosx)#

= #sin(2x-x)/(cos2xcosx)=sinx/(cos2xcosx)#

and #cot2x-cotx=(cos2x)/(sin2x)-cosx/sinx=(sinxcos2x-cosxsin2x)/(sin2xsinx)#

= #sin(x-2x)/(sin2xsinx)=-sinx/(sin2xsinx)#

Hence #1/(tan2x-tanx)-1/(cot2x-cotx)=1# can be written as

#(cos2xcosx)/sinx+(sin2xsinx)/sinx=1#

or #(cos2xcosx+sin2xsinx)/sinx=1#

or #cos(2x-x)/sinx=1#

or #cosx/sinx=1# i.e. #cotx=1=cot(pi/4)#

Hence #x=npi+pi/4#