$2 tan x - tan 2 x + 2 a = 1 - tan 2 x {tan}^{2} x$ $2tanx - tan2x + 2a = 1 - tan2xtan^2x$ What is a?

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Answer 1 · 2017-07-18T03:08:04

$a = \frac{1}{2}$ $a = 1/2$

$2 tan x - tan 2 x + 2 a = 1 - tan 2 x {tan}^{2} x$ $2 tan x - tan 2x + 2a = 1 - tan 2x tan^2 x$ $\to i$ $->i$

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

$\to tan 2 x (1 - {tan}^{2} x) = 2 tan x$ $-> tan 2x ( 1 - tan^2 x) = 2 tan x$

$\to tan 2 x - tan 2 x {tan}^{2} x = 2 tan x$ $-> tan 2x - tan 2x tan^2 x = 2 tan x$ ---> plug in #i

$cancel(tan 2x) - cancel(tan 2x tan^2 x) - cancel(tan 2x) + 2a = 1 - cancel(tan 2x tan^2 x)$

2a = 1

$a = 1/2$

2tanx - tan2x + 2a = 1 - tan2xtan^2x2tanx−tan2x+2a=1−tan2xtan2x2tanx - tan2x + 2a = 1 - tan2xtan^2x What is a?