Question

How can i solve this

tan2x/(tan4x-tan2x)​

Answer 1

the answer is yes i did this before so hope that helps

Answer 2

Answer: = cos 4x

`(tan2x)/(tan4x-tan2x)\\\\=(tan2x)/((2tan2x)/(1-tan^(2)2x )-tan2x )\\\\=(tan2x(1-tan^(2)2x) )/(2tan2x-tan2x(1-tan^(2)2x) )\\\\=(tan2x(1-tan^(2)2x) )/(tan2x(2-1+tan^(2)2x) )\\\\=(1-tan^(2)2x )/(1+tan^(2)2x ) \\\\=(1-(sin^(2)2x )/(cos^(2)2x ) )/(1+(sin^(2)2x )/(cos^(2)2x ) )\\\\=(cos^(2)2x-sin^(2)2x )/(cos^(2)2x+sin^(2)2x ) \\\\=(cos4x)/(1)\\\\=cos4x`

Explanation: