Question

$\rm \bold{ If \: f ( x ) = tan \: x , \: Then \: Show \: That \: f (2x) = \frac{2f(x)}{1 -[f(x)] } }$






⭐Thenku ~​

Answer 1

$\qquad \sf  \: \huge \bf༆ Answer ༄$

Here's the solution ~

$\qquad \sf  \dashrightarrow \: f(x) = \tan(x)$

$\qquad \sf  \dashrightarrow \: f(2x) = \tan(2x)$

$\qquad \sf  \dashrightarrow \: f(2x) = \dfrac{2 \tan(x) }{1 - { \tan {}^{2} (x) }^{} }$

replace tan(x) with f(x) in the given equation ~

$\qquad \sf  \dashrightarrow \: f(2x) = \dfrac{2 (f(x) )}{1 - { (f {}^{} (x)) {}^{2} }^{} }$

Answer 2

The expression that was proved is $tan2x=\frac{2[f(x)]}{1-[f(x)^2]}$

Double angle of function

Given the trigonometry identity

f(x) = tanx

f(2x)  tan2x

From double angle theorem;

$tan2x=\frac{tanx+tanx}{1-tanx(tanx)} \\tan2x=\frac{2tanx}{1-tan^2x}$

Since f(x) = tan(x), substitute into the result to have:

$tan2x=\frac{2[f(x)]}{1-[f(x)^2]}$ (Proved)

Learn more on double angle here: https://brainly.com/question/14691