# Verify the identity.\cot^2\theta-\cos^2\theta=\cot^2\theta\cos^\theta

Verify the identity.
$$\displaystyle{{\cot}^{{2}}\theta}-{{\cos}^{{2}}\theta}={{\cot}^{{2}}\theta}{{\cos}^{\theta}}$$

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Szeteib
Given:
$$\displaystyle{{\cot}^{{2}}\theta}-{{\cos}^{{2}}\theta}={{\cot}^{{2}}\theta}{{\cos}^{\theta}}$$
Manipulating right side
$$\displaystyle{{\cot}^{{2}}\theta}{{\cos}^{{2}}\theta}$$ Rewrite using trigonometry identities: $$\displaystyle{{\cos}^{{2}}\theta}={1}-{{\sin}^{{2}}\theta}$$
$$\displaystyle={\left({1}-{{\sin}^{{2}}\theta}\right)}{{\cot}^{{2}}\theta}$$
Expand $$\displaystyle{\left({1}-{{\sin}^{{2}}\theta}\right)}{{\cot}^{{2}}\theta}:{{\cot}^{{2}}\theta}-{{\cot}^{{2}}\theta}{{\sin}^{{2}}\theta}$$
$$\displaystyle={{\cot}^{{2}}\theta}-{{\cot}^{{2}}\theta}{{\sin}^{{2}}\theta}$$
Simplify using: $$\displaystyle{{\cot}^{{2}}\theta}={\frac{{{{\cos}^{{2}}\theta}}}{{{{\sin}^{{2}}\theta}}}}$$
$$\displaystyle={{\cot}^{{2}}\theta}-{\frac{{{{\cos}^{{2}}\theta}}}{{{\sin}^{\theta}}}}\times{{\sin}^{{2}}\theta}$$
$$\displaystyle={{\cot}^{{2}}\theta}-{{\cos}^{\theta}}$$
We showed that the two sides could take the same form.
Hence proved
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