Given: sec x = -2

Given that: \(\pi<x<\frac{3\pi}{2}\)
So, x lies in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.
Now, we know that
cos x = \(\frac{1}{sec\,x}\)
Putting the values, we get
cos x = \(\frac{1}{-2}\) .....(i)
We know that,
cos2 x + sin2 x = 1
Putting the values, we get

Since, x in IIIrd quadrant and sinx is negative in IIIrd quadrant
∴ sin x = \(-\frac{\sqrt{3}}{2}\)
Now,
tan x = \(\frac{sin\,x}{cos\,x}\)
Putting the values, we get

Hence, the values of other trigonometric Functions are:
