How do you simplify $\frac{2 tan (x + 1)}{1 - {tan}^{2} (x + 1)}$ $(2tan(x+1))/(1-tan^2(x+1))$ ?

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Answer 1 · 2015-08-16T16:02:07

$\frac{2 tan (x + 1)}{1 - {tan}^{2} (x + 1)} = tan 2 (x + 1)$ $(2tan (x+1))/(1-tan^2 (x+1)) = tan 2(x+1)$

$tan 2 A = \frac{2 tan A}{1 - {tan}^{2} A}$ $tan 2A = (2tan A)/(1-tan^2 A)$

$\frac{2 tan (x + 1)}{1 - {tan}^{2} (x + 1)} = tan 2 (x + 1)$ $(2tan (x+1))/(1-tan^2 (x+1)) = tan 2(x+1)$

How do you simplify (2tan(x+1))/(1-tan^2(x+1))2tan(x+1)1−tan2(x+1)(2tan(x+1))/(1-tan^2(x+1))?