## Trigonometry (11th Edition) Clone

There are 2 possible values of $$\frac{\sin x+\cos x}{\sec x}$$ whether $$\frac{\sin x+\cos x}{\sec x}=\frac{8-2\sqrt2}{9}$$ or $$\frac{\sin x+\cos x}{\sec x}=\frac{8+2\sqrt2}{9}$$
$$\csc x=-3\hspace{2cm}A=\frac{\sin x+\cos x}{\sec x}$$ Since it is hard to represent trigonometric expressions in terms of $\csc x$, it is better if we change it first into $\sin x$. - Reciprocal Identity: $$\csc x=\frac{1}{\sin x}$$ $$\sin x=\frac{1}{\csc x}=\frac{1}{-3}=-\frac{1}{3}$$ - Pythagorean Identity: $$\cos^2 x=1-\sin^2 x=1-(-\frac{1}{3})^2=1-\frac{1}{9}=\frac{8}{9}$$ $$\cos x=\pm\frac{\sqrt 8}{3}=\pm\frac{2\sqrt 2}{3}$$ Now we try to represent $A$ in terms of $\sin x$ and $\cos x$. - Reciprocal Identity: $$\sec x=\frac{1}{\cos x}$$ Therefore, $$A=\frac{\sin x+\cos x}{\frac{1}{\cos x}}$$ $$A=\cos x(\sin x+\cos x)$$ Now we replace the values of $\sin x$ and $\cos x$ into $A$. Remember that there are 2 possible values of $\cos x$. 1) As $\sin x=-\frac{1}{3}$ and $\cos x=\frac{2\sqrt2}{3}$ $$A=\frac{2\sqrt2}{3}(-\frac{1}{3}+\frac{2\sqrt2}{3})$$ $$A=\frac{2\sqrt2}{3}\times\frac{2\sqrt2-1}{3}$$ $$A=\frac{2\sqrt2(2\sqrt2-1)}{9}$$ $$A=\frac{8-2\sqrt2}{9}$$ 2) As $\sin x=-\frac{1}{3}$ and $\cos x=-\frac{2\sqrt2}{3}$ $$A=-\frac{2\sqrt2}{3}(-\frac{1}{3}-\frac{2\sqrt2}{3})$$ $$A=-\frac{2\sqrt2}{3}\times\frac{(-2\sqrt2-1)}{3}$$ $$A=\frac{-2\sqrt2(-2\sqrt2-1)}{9}$$ $$A=\frac{8+2\sqrt2}{9}$$