## Trigonometry (11th Edition) Clone

$$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sin x}$$
According to Pythagorean Identities, $$\cot^2x+1=\csc^2 x\hspace{1.5cm}(1)$$ Also, according to Reciprocal Identities, $$\csc x=\frac{1}{\sin x}$$ So, $$\csc^2 x=\frac{1}{\sin^2 x}\hspace{1.5cm}(2)$$ We can combine $(1)$ and $(2)$ together and have $$\cot^2 x+1=\frac{1}{\sin^2 x}$$ $$\cot^2 x=\frac{1}{\sin^2 x}-1$$ $$\cot^2 x=\frac{1-\sin^2 x}{\sin^2 x}$$ Now we take the square root of both sides: $$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sqrt{\sin^2 x}}$$ (Do not forget the $\pm$ sign) $$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{|\sin x|}$$ Since we already have the $\pm$ sign, we can eliminate the absolute value sign. $$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sin x}$$