How do you prove $[\frac{2 tan x}{1 - {tan}^{2} x}] + [\frac{1}{2 {cos}^{2} x - 1}] = \frac{cos x + sin x}{cos x - sin x}$ $[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)$ ?

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How do you prove $[\frac{2 tan x}{1 - {tan}^{2} x}] + [\frac{1}{2 {cos}^{2} x - 1}] = \frac{cos x + sin x}{cos x - sin x}$ $[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)$ ?

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Answer 1 · 2018-03-26T13:55:33

Please see below.

Explanation:

We know that,

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

$(2) cos 2 x = 2 {cos}^{2} x - 1 = {cos}^{2} x - {sin}^{2} x$ $color(red)((2)cos2x=2cos^2x-1=cos^2x-sin^2x$

$(3) sin 2 x = 2 sin x cos x, and {cos}^{2} x + {sin}^{2} x = 1$ $color(red)((3)sin2x=2sinxcosx,and cos^2x+sin^2x=1$

We have,

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

Now,

$L H S = \frac{2 tan x}{1 - {tan}^{2} x} + \frac{1}{2 {cos}^{2} x - 1}$ $LHS=(2tanx)/(1-tan^2x)+1/(2cos^2x-1$

$=tan2x+1/(cos2x).......to$ Using $(1) and (2)$

$=(sin2x)/(cos2x)+1/(cos2x)$

$=(1+sin2x)/(cos2x)$

$=(sin^2x+cos^2x+2sinxcosx)/(cos^2x-sin^2x).....to$ Using $(2) and (3)$

$=(cosx+sinx)^cancel(2)/(cancel((cosx+sinx))(cosx-sinx))$

$=(cosx+sinx)/(cosx-sinx)$

$=RHS$

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How do you prove $[\frac{2 tan x}{1 - {tan}^{2} x}] + [\frac{1}{2 {cos}^{2} x - 1}] = \frac{cos x + sin x}{cos x - sin x}$ $[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)$ ?

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Explanation:

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Explanation:

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How do you prove $[\frac{2 tan x}{1 - {tan}^{2} x}] + [\frac{1}{2 {cos}^{2} x - 1}] = \frac{cos x + sin x}{cos x - sin x}$ $[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)$ ?