((tan2x)(1+tanx))/tanx = 2/(1-tanx)

Question

can u please help me to prove this

rhs=lhs

Answer 1

[(tan2x)(1+tanx)]/tanx=2/(1-tanx)

working with the numerator of the left side only...(I haven't forgotten the denominator tanx)

tan2x=2tanx/(1-tan²x)

(2tanx/(1-tan²x)*(1+tanx)

[2tanx/(1-tanx)(1+tanx)]*(1+tanx)

(2tanx)(1+tanx)/(1-tanx)(1+tanx) ((1+tanx) is in both the numerator and denominator and therefore cancel out)

2tanx/(1-tanx)

now bring in the denominator tanx

2tanx/(1-tanx) "divided by" tanx

(2tanx/(1-tanx)(1/tanx)

you have tanx in the numerator of the first fraction and in the denominator of the second fraction

therefore they cancel out leaving you with...

2/(1-tanx) which is the right side and you had to show that the left side equals the right side, or, in other words,

you take the left side and use other trig identities to change the left side into the right side