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zzz1090 zzz1090
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11 years ago
How do you prove the trigonometric identity below?

tan3x - tanx                2tanx
------------------   =   --------------------
1+tan3xtanx        1-(tanx)(tanx)

I know that the right side is equal to tan2x, but how do you prove the left side to be equal?
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#1
11 years ago
using tan (A-B) = (tan A - tan B)/(1+ tan A tan B)

and A = 3x and B = x

we get tan 2x = (tan 3x-tan x)/(1+tan 3x tan x)

so LHS = tan 2x
as RHs is also tan 2x both are same

math kp
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#2
11 years ago
(tan3x - tanx)/(1+tan3xtanx)
= tan(3x-x)
= tan(2x)
= 2tanx/[1-(tanx)(tanx)]
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#3
11 years ago
= [tan3x - tanx]/[1+tan3xtanx]
= tan (3x - x)
= tan 2x
= tan (x + x)
= [2tanx]/[1-(tanx)(tanx)]
then
[tan3x - tanx]/[1+tan3xtanx] = [2tanx]/[1-(tanx)(tanx)]
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tomanaditomanadi
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#4
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11 years ago
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#5
11 years ago
(tan 3x - tan x)/(1 + tan 3x tan x) = (2 tan x)/[1 - (tan x)(tan x)]

LHS
= (tan 3x - tan x)/(1 + tan 3x tan x)
= tan (3x - x)
= tan 2x
= tan (x + x)
= (2 tan x)/[1 - (tan x)(tan x)]
= RHS
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