How do you prove #(tan3t-tant)/(1+tan3t*tant)=(2tant)/(1-tan^2t)#?

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Answer 1 · 2016-10-16T09:47:01

#LHS=(tan3t-tant)/(1+tan3t*tant)#

#=((sin3t)/(cos3t)-(sint)/(cost))/(1+(sin3t)/(cos3t)*(sint)/(cost))#

#=(sin3tcost-cos3tsint )/(cos3tcost+sin3tsint)#

#=(sin2t)/(cos2t)#

#=tan2t#

#=(2tant)/(1-tan^2t)=RHS#

Proved