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

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