How do you prove $\frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} = 1 + tan 2 x . tan x$ $(1+tan^2x)/(1-tan^2x)=1+tan2x.tanx$ ?

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How do you prove $\frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} = 1 + tan 2 x . tan x$ $(1+tan^2x)/(1-tan^2x)=1+tan2x.tanx$ ?

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Answer 1 · 2018-05-01T15:03:46

Check the explanation below

Explanation:

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

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

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

$tan x = \frac{sin x}{cos x}$ $color(red)(tanx=sinx/cosx$

$= 1 + \frac{2 \frac{{sin}^{2} x}{{cos}^{2} x}}{1 - \frac{{sin}^{2} x}{{cos}^{2} x}}$ $=1+(2color(green)(sin^2x/cos^2x))/(1-color(green)(sin^2x/cos^2x)$

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

Using the Double Angle Identities

${cos}^{2} x - {sin}^{2} x = cos 2 x$ $color(red)(cos^2x-sin^2x=cos2x)$
$2 sin x cos x = sin 2 x$ $color(red)(2sinxcosx=sin2x$

$= 1 + \frac{2 {sin}^{2} x}{cos 2 x}$ $=1+(2sin^2x)/(cos2x)$

$= 1 + 2 \frac{{sin}^{2} x}{cos 2 x} \times \frac{cos x}{cos x}$ $=1+2sin^2x/(cos2x)color(green)(xxcosx/cosx$

$= 1 + \frac{sin x \cdot (2 sin x cos x)}{cos x \cdot cos 2 x}$ $=1+(sinx*(2sinxcosx))/(cosx*cos2x)$

$= 1 + \frac{sin x \cdot sin 2 x}{cos x \cdot cos 2 x}$ $=1+color(green)(sinx*color(blue)(sin2x))/(color(green)(cosx*color(blue)(cos2x)$

$= 1 + tan 2 x \cdot tan x$ $=1+tan2x*tanx$ $\to R.T.P$ $color(green)(rarr " R.T.P"$

I hope this was helpful :)

Answer 2 · 2018-05-01T18:06:57

We seek to prove the identity:

$\frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} \equiv 1 + tan 2 x tan x$ $(1+tan^2x)/(1-tan^2x) -=1+tan2xtanx$

We can utilise the tangent double angle formula:

$tan 2 A \equiv \frac{2 tan A}{1 - {tan}^{2} A}$ $tan2A -= (2tanA)/(1-tan^2A)$

Consider the RHS:

$R H S = 1 + tan 2 x tan x$ $RHS = 1+tan2xtanx$

$\ \ \ \ \ \ \ \ = 1+((2tanx)/(1-tan^2x))tanx$

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

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

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

$\ \ \ \ \ \ \ \ = LHS \ \ \$ QED

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How do you prove $\frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} = 1 + tan 2 x . tan x$ $(1+tan^2x)/(1-tan^2x)=1+tan2x.tanx$ ?

2 Answers

Explanation:

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Explanation:

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How do you prove $\frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} = 1 + tan 2 x . tan x$ $(1+tan^2x)/(1-tan^2x)=1+tan2x.tanx$ ?