<s>[INST] If $\sec x=\frac{a}{b}$, find the exact value of $\cot x-\frac1{\sin x}$, for positive $a$ and $b$, and for $x\in(\frac{3\pi}{2},2\pi)$ 
If $\sec(x) = \frac{a}{b}$, find the exact value of
$$\cot(x) - \frac{1}{\sin(x)}$$ and $a$, $b$ are positive real values and $x\in(\frac{3\pi}{2},2\pi)$


*

*I know that $\sec(x) = \frac{1}{\cos(x)}, $ so:  $\cos(x) = \frac{b}{a}$

*I used the right angled triangle to find the third side, which is $\sqrt{a^2-b^2}$

*I used reciprocal functions of $\tan$ and $\sin$ to obtain the values:

$cosec(x) = \frac{a}{(\sqrt{a^2-b^2})}$ and $\cot(x) = \frac{b}{\sqrt{a^2-b^2}}$
and finally getting $\frac{(b-a)}{\sqrt{a^2-b^2}}$ as my answer. But that is not correct.
The correct answer is: $\frac{\sqrt{a^2-b^2}}{a+b}$. Now I am wondering how.
 [/INST] *

*Remember, when using a right-angled triangle to determine
trigonometric values, we are actually just determining their
absolute values (i.e., dropping any negative sign).
This is evidenced by the fact that the process evaluates ratios of triangle lengths, each of which is expressed as a nonnegative value in the form $\sqrt{u^2\pm v^2}.$
In the given example, since $x$ is located in the fourth quadrant, $\operatorname{cosec}x$ must be negative and so equals $\displaystyle\mathbf-\left(\frac a{\sqrt{a^2-b^2}}\right)$ rather than $\displaystyle\frac a{\sqrt{a^2-b^2}}.$
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