## Trigonometry (11th Edition) Clone

$$\cos2x=\frac{2-\sec^2x}{\sec^2x}$$ The equation is verified to be an identity as below.
$$\cos2x=\frac{2-\sec^2x}{\sec^2x}$$ In this exercise, we would tackle the right side first. $$X=\frac{2-\sec^2x}{\sec^2x}$$ - For $\sec x$, we rewrite according to the following reciprocal identity: $$\sec x=\frac{1}{\cos x}$$ Apply to $X$: $$X=\frac{2-\frac{1}{\cos^2x}}{\frac{1}{\cos^2x}}$$ $$X=\frac{\frac{2\cos^2x-1}{\cos^2x}}{\frac{1}{\cos^2x}}$$ $$X=\frac{2\cos^2x-1}{1}$$ $$X=2\cos^2x-1$$ - Now recall that $2\cos^2x-1=\cos2x$. Therefore, $$X=\cos2x$$ That means $$\cos2x=\frac{2-\sec^2x}{\sec^2x}$$ As 2 sides are equal, the equation is an identity.