If tan x = -12/5, where x lies in the second quadrant, what is the value of Sin x-cot x ?

$$\tan x = -\frac{12}{5}$$​, x lies in the second quadrant,

In the second quadrant, $$\sin x$$​ is positive, and $$\cos x$$​ is negative.

The trigonometric identity: $$\tan x = \frac{\sin x}{\cos x}$$​​

and $$\sin^2 x + \cos^2 x = 1$$​​

​$$\tan x = \frac{\sin x}{\cos x}$$​​

​$$\sin x = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$$​​

Solution:  

$$\because$$ In the second quadrant, $$\sin x$$​​is positive, and $$\cos x$$​is negative. 

So, 

​$$\sin x - \cot x = \frac{12}{13} -\left(-\frac{5}{12}\right) \\ \ \\=\frac{144 -(-65)}{156} \\ \ \\=\bf \frac{209}{156}$$ 

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