Answer
There are 2 possible values for $$\frac{\sec x-\tan x}{\sin x}$$, which is $$\frac{\sec x-\tan x}{\sin x}=\frac{25\sqrt6-60}{12}$$ or $$\frac{\sec x-\tan x}{\sin x}=\frac{-25\sqrt6-60}{12}$$
Work Step by Step
$$\cos x=\frac{1}{5}\hspace{2cm}A=\frac{\sec x-\tan x}{\sin x}$$
To find values of $A$, we need to represent $A$ first in terms of $\cos x$ as much as we can.
- Reciprocal Identity:
$$\sec x=\frac{1}{\cos x}$$
- Quotient Identity:
$$\tan x=\frac{\sin x}{\cos x}$$
Therefore, $$A=\frac{\frac{1}{\cos x}-\frac{\sin x}{\cos x}}{\sin x}$$
$$A=\frac{\frac{1-\sin x}{\cos x}}{\sin x}$$
$$A=\frac{1-\sin x}{\sin x\cos x}$$
So we have represented $A$ into $\sin x$ and $\cos x$. And we can find $\sin x$ according to $\cos x$.
- Pythagorean Identity:
$$\sin^2 x=1-\cos^2 x=1-(\frac{1}{5})^2=1-\frac{1}{25}=\frac{24}{25}$$
$$\sin x=\pm\frac{\sqrt{24}}{5}=\pm\frac{2\sqrt6}{5}$$
1) As $\sin x=\frac{2\sqrt6}{5}$ and $\cos x=\frac{1}{5}$
$$A=\frac{1-\frac{2\sqrt6}{5}}{\frac{2\sqrt6}{5}\times\frac{1}{5}}$$
$$A=\frac{\frac{5-2\sqrt6}{5}}{\frac{2\sqrt6}{25}}$$
$$A=\frac{25(5-2\sqrt6)}{5\times2\sqrt6}$$
$$A=\frac{5(5-2\sqrt6)}{2\sqrt6}$$
$$A=\frac{5\sqrt6(5-2\sqrt6)}{2\times6}$$ (multiply both numerator and denominator by $\sqrt6$)
$$A=\frac{25\sqrt6-60}{12}$$
2) As $\sin x=-\frac{2\sqrt6}{5}$ and $\cos x=\frac{1}{5}$
$$A=\frac{1-(-\frac{2\sqrt6}{5})}{(-\frac{2\sqrt6}{5})\times\frac{1}{5}}$$
$$A=\frac{1+\frac{2\sqrt6}{5}}{-\frac{2\sqrt6}{25}}$$
$$A=-\frac{\frac{5+2\sqrt6}{5}}{\frac{2\sqrt6}{25}}$$
$$A=-\frac{25(5+2\sqrt6)}{5\times2\sqrt6}$$
$$A=-\frac{5(5+2\sqrt6)}{2\sqrt6}$$
$$A=-\frac{5\sqrt6(5+2\sqrt6)}{2\times6}$$ (multiply both numerator and denominator by $\sqrt6$)
$$A=\frac{-25\sqrt6-60}{12}$$
