# How do you prove (1+tan^2x)/(1-tan^2x)=1+tan2x.tanx ?

May 1, 2018

Check the explanation below

#### Explanation:

$\frac{1 + {\tan}^{2} x}{1 - {\tan}^{2} x}$

$= \frac{1 \textcolor{g r e e n}{- {\tan}^{2} x + {\tan}^{2} x} + {\tan}^{2} x}{1 - {\tan}^{2} x}$

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

color(red)(tanx=sinx/cosx

=1+(2color(green)(sin^2x/cos^2x))/(1-color(green)(sin^2x/cos^2x)

$= 1 + \frac{2 {\sin}^{2} x}{{\cos}^{2} x - {\sin}^{2} x}$

Using the Double Angle Identities

• $\textcolor{red}{{\cos}^{2} x - {\sin}^{2} x = \cos 2 x}$
• color(red)(2sinxcosx=sin2x

$= 1 + \frac{2 {\sin}^{2} x}{\cos 2 x}$

=1+2sin^2x/(cos2x)color(green)(xxcosx/cosx

$= 1 + \frac{\sin x \cdot \left(2 \sin x \cos x\right)}{\cos x \cdot \cos 2 x}$

=1+color(green)(sinx*color(blue)(sin2x))/(color(green)(cosx*color(blue)(cos2x)

$= 1 + \tan 2 x \cdot \tan x$color(green)(rarr " R.T.P"

I hope this was helpful :)

May 1, 2018

We seek to prove the identity:

$\frac{1 + {\tan}^{2} x}{1 - {\tan}^{2} x} \equiv 1 + \tan 2 x \tan x$

We can utilise the tangent double angle formula:

$\tan 2 A \equiv \frac{2 \tan A}{1 - {\tan}^{2} A}$

Consider the RHS:

$R H S = 1 + \tan 2 x \tan x$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 + \left(\frac{2 \tan x}{1 - {\tan}^{2} x}\right) \tan x$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 + \frac{2 {\tan}^{2} x}{1 - {\tan}^{2} x}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left(1 - {\tan}^{2} x\right) + \left(2 {\tan}^{2} x\right)}{1 - {\tan}^{2} x}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1 + {\tan}^{2} x}{1 - {\tan}^{2} x}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = L H S \setminus \setminus \setminus$ QED