Question

if x is an acute angle such that tanx + cotx = 2 find
$tan {}^{6}x + cot {}^{6}x$

Answer 1

Given : tan x + cot x = 2.

It can be written as:

⇒ tanx + (1/tanx) = 2

⇒ tan^2x + 1 = 2 tanx

⇒ tan^2x + 1 - 2tanx = 0

⇒ (tan x)^2 + (1)^2 - 2(tanx)(1) = 0

It is in the form of a^2 + b^2 - 2ab = (a - b)^2

⇒ (tanx - 1)^2 = 0

⇒ tan x = 1

⇒ tan x = tan 45

⇒ x = 45.

Now,

⇒ tan^6x + cot^6x

⇒ (tan x)^6 + (cot x)^6

⇒ 1 + 1

⇒ 2.

Therefore, tan^6x + cot^6x = 2.

Hope this helps!

Answer 2

since, x is an acute angle ,x < 90°

tanx + cotx = 2

tan x + 1 / tan x = 2

tan^2 x + 1 = 2 tan x

tan^2 x + 1 - 2 tan x = 0

( tan x )^2 + ( 1 )^2 - 2 ( tan x ) ( 1 ) = 0

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use identiy : a^2 + b^2 - 2ab = ( a - b )^2
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( tan x - 1 )^2 = 0

tan x - 1 = 0

tan x = 1

x = 45°

therefore ,

tan^6 x + cot^6 x = tan^6 45° + cot^6 45°

= (1 )^6 + (1 )^6 =2
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