Question

Solve:
$\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}$

• A. $-\tan 2x$
• B. $\sin 2x$
• C. $-\cos 2x$
• D. None of these

Answer 1

The correct option is C $-\cos 2x$

$\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}$

$\Rightarrow \frac{({\sin }^{4}x{)}^2-({\cos }^{4}x{)}^2}{({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x}{x}^2-y^2=(x+y)(x-y)$

$\Rightarrow \frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{4}x-{\cos }^{4}x)}{{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x}$

$\Rightarrow \frac{({\sin }^{4}x+{\cos }^{4}x)({\sin }^{2}x{)}^2-({\cos }^{2}x{)}^2}{({\sin }^{4}x+{\cos }^{4}x)}$

$\Rightarrow ({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)[x^2-y^2=(x+y\left)\right(x-y\left)\right]$

$\Rightarrow \left(1\right)(-({\cos }^{2}x-{\sin }^{2}x\left)\right)$

$\Rightarrow -\left(\cos 2x\right)$

$\therefore \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}=-\cos \left(2x\right)$