Answer
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 $$\frac{\sin x+\cos x}{\sec x}=\frac{8+2\sqrt2}{9}$$
Work Step by Step
$$\csc x=-3\hspace{2cm}A=\frac{\sin x+\cos x}{\sec x}$$
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 $\cos x=-\frac{2\sqrt2}{3}$
$$A=-\frac{2\sqrt2}{3}(-\frac{1}{3}-\frac{2\sqrt2}{3})$$
$$A=-\frac{2\sqrt2}{3}\times\frac{(-2\sqrt2-1)}{3}$$
$$A=\frac{-2\sqrt2(-2\sqrt2-1)}{9}$$
$$A=\frac{8+2\sqrt2}{9}$$
