If the function $f(x)=\sin x-\cos ^2 x$ is defined on the interval $[-\pi, \pi]$, then $f$ is strictly increasing in the interval

• A

  $\left(\frac{-5 \pi}{6}, \frac{-\pi}{6}\right) \cup\left(\frac{-\pi}{6}, \frac{\pi}{2}\right)$
• B

  $\left(\frac{-\pi}{2}, \frac{-\pi}{6}\right)$
• C

  $\left(\frac{-5 \pi}{6}, \frac{\pi}{2}\right)$
• Correct answer
  D

  $\left(\frac{-5 \pi}{6}, \frac{-\pi}{2}\right) \cup\left(\frac{-\pi}{6}, \frac{\pi}{2}\right)$

Hence, the correct answer is option: 'D'.