Hi,

The question I have is:

Given that $\displaystyle \sec \theta = -2$, $\displaystyle \cot \theta = -1/3\sqrt{3}$ and $\displaystyle -\pi < 0 < \pi$, find the exact value of the angle $\displaystyle \theta$ in radians. Justify your answer.

Is the following along the right lines?

$\displaystyle \sec\theta=1/cos\theta$ therefore $\displaystyle \cos\theta=1/\sec \theta=-1/2$

$\displaystyle \cot\theta=1/\tan\theta$ therefore $\displaystyle \tan\theta=1/\cot\theta=-3 \sqrt{3}/3$

$\displaystyle \sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}$

Since both $\displaystyle \tan\theta$ and $\displaystyle \cos\theta$ are negative and lie in the range $\displaystyle -\pi < 0 < \pi$, then $\displaystyle \theta$ must lie in the 2nd quadrant so is positive.

Thanks for your help