Question

How can i solve thistan2x/(tan4x-tan2x)​

Answer 1

Answer: = cos 4x

$\frac{tan2x}{tan4x-tan2x}\\\\=\frac{tan2x}{\frac{2tan2x}{1-tan^{2}2x }-tan2x }\\\\=\frac{tan2x(1-tan^{2}2x) }{2tan2x-tan2x(1-tan^{2}2x) }\\\\=\frac{tan2x(1-tan^{2}2x) }{tan2x(2-1+tan^{2}2x) }\\\\=\frac{1-tan^{2}2x }{1+tan^{2}2x } \\\\=\frac{1-\frac{sin^{2}2x }{cos^{2}2x } }{1+\frac{sin^{2}2x }{cos^{2}2x } }\\\\=\frac{cos^{2}2x-sin^{2}2x }{cos^{2}2x+sin^{2}2x } \\\\=\frac{cos4x}{1}\\\\=cos4x$

Step-by-step explanation:

Answer 2

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