# Verify the identify.\frac{\csc^2\theta}{1+\tan^2\theta}=\cot^2\theta

Verify the identify.
$$\displaystyle{\frac{{{{\csc}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}={{\cot}^{{2}}\theta}$$

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Mayme
We know that
$$\displaystyle{\cot{{x}}}={\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}$$
$$\displaystyle{\tan{{x}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}$$
$$\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}$$
$$\displaystyle{L}{H}{S}={\frac{{{{\csc}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}$$
$$\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{1}+{\frac{{{{\sin}^{{2}}\theta}}}{{{{\cos}^{{2}}\theta}}}}}}}$$
$$\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{\frac{{{{\cos}^{{2}}\theta}+{{\sin}^{{2}}\theta}}}{{{{\cos}^{{2}}\theta}}}}}}}$$
$$\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{\frac{{{1}}}{{{{\cos}^{{2}}\theta}}}}}}}$$
$$\displaystyle={\left({\frac{{{\cos{\theta}}}}{{{\sin{\theta}}}}}\right)}^{{2}}$$
$$\displaystyle={{\cot}^{{2}}\theta}$$
$$\displaystyle={R}{H}{S}$$