How do you prove #(1+tanx) tan2x = (2tanx)/(1-tanx)#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Konstantinos Michailidis Mar 7, 2016 From the identity #tan(A+B)=[tanA+tanB]/[1-tanA*tanB]# for #A=B=x# we have that #tan(x+x)=2*tanx/(1-tan^2x)=> tan2x=[2*tanx]/[(1-tanx)*(1+tanx)]=> (1+tanx)*tan2x=[2*tanx]/(1-tanx)# Answer link Related questions What are Double Angle Identities? How do you use a double angle identity to find the exact value of each expression? How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#? How do you find all solutions for #4sinthetacostheta=sqrt(3)# for the interval #[0,2pi]#? How do you simplify #cosx(2sinx + cosx)-sin^2x#? If #tan x = 0.3#, then how do you find tan 2x? If #sin x= 5/3#, what is the sin 2x equal to? How do you prove #cos2A = 2cos^2 A - 1#? See all questions in Double Angle Identities Impact of this question 6273 views around the world You can reuse this answer Creative Commons License