Trigonometry (11th Edition) Clone

$$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$ The equation is an identity as proved below.
$$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$ The left side would be analyzed first. $$X=\csc^4x-\cot^4x$$ $$X=(\csc^2x-\cot^2x)(\csc^2x+\cot^2x)$$ - $\csc x$ and $\cot x$ would be rewritten according to the following identities: $$\csc x=\frac{1}{\sin x}\hspace{2cm}\cot x=\frac{\cos x}{\sin x}$$ Therefore, $$\csc^2x=\frac{1}{\sin^2x}\hspace{2cm}\cot^2x=\frac{\cos^2x}{\sin^2x}$$ Apply them into $X$: $$X=\Big(\frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}\Big)\Big(\frac{1}{\sin^2x}+\frac{\cos^2x}{\sin^2x}\Big)$$ $$X=\frac{1-\cos^2x}{\sin^2x}\times\frac{1+\cos^2x}{\sin^2x}$$ $$X=\frac{(1-\cos^2x)(1+\cos^2x)}{\sin^4x}$$ - We can rewrite $(1-\cos^2x)$ on the numerator into $\sin^2x$ according to Pythagorean identity. $$X=\frac{\sin^2x(1+\cos^2x)}{\sin^4x}$$ $$X=\frac{1+\cos^2x}{\sin^2x}$$ - Finally, also from Pythagorean identity, $\sin^2x$ in the denominator can be written into $(1-\cos^2x)$ $$X=\frac{1+\cos^2x}{1-\cos^2x}$$ So we can conclude that $$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$ The equation is therefore an identity.