Find the solution: \frac{1+\tan(x)+\tan^2(x)+...+\tan^n(x)+...}{1-\tan(x)+\tan^2(x)-...+(-1)^n\tan^n(x)...}=1+\sin(2x)

Brooklyn1wp

Brooklyn1wp

Answered question

2022-02-26

Find the solution:
1+tan(x)+tan2(x)++tann(x)+1tan(x)+tan2(x)+(1)ntann(x)=1+sin(2x)

Answer & Explanation

Abbey Hope

Abbey Hope

Beginner2022-02-27Added 7 answers

If |tanx|1 both numerator and denominator of the left-hand side are not finite, because the geometric series
N=1+tan(x)+tan2(x)++tann(x)+
andD=1tan(x)+tan2(x)+(1)ntann(x)+
don't converge. If |tanx|<1, we have
N=limntannx1tanx1=01tanx1=11tanx
and
D=limn(tanx)n1(tanx)1=01tanx1=11+tanx
and the given equation is equivalent to
1+tanx1tanx=1+2tanx1+tan2x
1+tanx1tanx=(1+tan2x)21+tan2x, |tanx|<1 (1)
because
sin2x=2tanx1+tan2x
To solve equation (1) we can observe that it is equivalent to
(1+tanx)(1+tan2x)(1tanx)(1+tan2x)=(1+tanx)2(1tanx)(1tanx)(1+tan2x)
|tanx|<1
and to
(1+tanx)(1+tan2x)=(1+tanx)2(1tanx)

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