Question

cot2x÷cot2x - cot4x + tan2x ÷ tan2x - tan4x​

Answer 1

Given:

$\bold{\cot^2x\div \cot^2x - \cot^4x + \tan^2x \div \tan^2x - \tan^4x}$

To find:

Simplify

Solution:

$\to \cot^2x\div \cot^2x - \cot^4x + \tan^2x \div \tan^2x - \tan^4x\\\\\to \frac{\cot^2x}{\cot^2x} - \cot^4x+\frac{\tan^2x}{\tan^2x}-\tan^4x\\\\\to 1 - \cot^4x+1-\tan^4x\\\\\to 2 - \frac{\cos^4 x}{\sin^4 x}- \frac{\sin^4 x}{\cos^4 x} \\\\\to \frac{2\sin^4 x \cos^4 x- \cos^8 x-\sin^8 x}{\sin^4 x \cos^4 x } \\\\\to \frac{- (-2\sin^4 x \cos^4 x+ \cos^8 x+\sin^8 x}{\sin^4 x \cos^4 x } \\\\\to \boxed{- \frac{(\sin^4x-\cos^4x)^2}{\sin^4 x \cos^4 x}}$