## Trigonometry (11th Edition) Clone

$$\csc x=\pm\frac{\sqrt{1-\cos^2 x}}{1-\cos^2 x}$$
From Reciprocal Identities: $$\csc x=\frac{1}{\sin x}\hspace{1.5cm}(1)$$ We see that we find a connection between $\csc x$ and $\sin x$. However, the exercise asks for a connection between $\csc x$ and $\cos x$. What we can do is find a way to represent $\sin x$ in terms of $\cos x$ From Pythagorean Identities, we also have: $$\sin^2 x+\cos^2 x=1$$ which means $$\sin^2 x=1-\cos^2 x$$ $$\sin x=\pm\sqrt{1-\cos^2 x}\hspace{1.5cm}(2)$$ (as we take the square root of both sides) As we found the way to represent $\sin x$ in terms of $\cos x$, we can adopt it for $\csc x$ in $(1)$: $$\csc x=\frac{1}{\pm\sqrt{1-\cos^2 x}}$$ $$\csc x=\pm\frac{1}{\sqrt{1-\cos^2 x}}$$ We rationalize the denominator by multiplying both numerator and denominator by $\sqrt{1-\cos^2 x}$ $$\csc x=\pm\frac{1}{\sqrt{1-\cos^2 x}}\frac{\sqrt{1-\cos^2 x}}{\sqrt{1-\cos^2 x}}$$ $$\csc x=\pm\frac{\sqrt{1-\cos^2 x}}{1-\cos^2 x}$$