Given: cos θ = \(\frac{-\sqrt{3}}2{}\)
Since, θ is in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.
We know that,
cos2 θ + sin2 θ = 1
Putting the values, we get
Since, θ in IIIrd quadrant and sinθ is negative in IIIrd quadrant
∴ sin θ = \(-\frac{1}{2}\)
Now,
tan θ = \(\frac{sin\,\theta}{cos\,\theta}\)
Putting the values, we get
= \(\sqrt{3}\)
Hence, the values of other trigonometric Functions are:
