Trigonometry (11th Edition) Clone

$$\frac{\cos x}{\sec x}+\frac{\sin x}{\csc x}=1$$
$$A=\frac{\cos x}{\sec x}+\frac{\sin x}{\csc x}$$ - Reciprocal Identities: $$\csc x=\frac{1}{\sin x}\hspace{2cm}\sec x=\frac{1}{\cos x}$$ So, $$\frac{\cos x}{\sec x}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x$$ and $$\frac{\sin x}{\csc x}=\frac{\sin x}{\frac{1}{\sin x}}=\sin^2 x$$ Apply these back to $A$: $$A=\cos^2x+\sin^2 x$$ $$A=1$$ (Pythagorean Identity)