Simplify This Trigonometric Identity

Question

$$\frac{\cos^2B-\sin^2B}{\sin^4B-\cos^4B}$$ Thanks in advance

Answer 1

Fill in details and justifications:

$$\frac{\cos^2x-\sin^2x}{\sin^4x-\cos^4x}=\frac{\cos^2x-\sin^2x}{(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)}=-1$$

Answer 2

HINT: use that $$\sin^4(B)-\cos^4(B)=(\sin^2(B)+\cos^2(B))(\sin^2(B)-\cos^2(B))$$

Answer 3

Hint:

$$\frac{\cos^2B-\sin^2B}{\sin^4B-\cos^4B}=\frac{\cos^2B-\sin^2B}{(\sin^2B+\cos^2B)(\sin^2B-\cos^2B)}=?$$

And

$$\sin x+\cos x=\sin x + \sin(90-x)$$