Given $\tan A = \sqrt3$ and $\csc B = -\sqrt2$ where $A,B \in[\pi,\frac{3}{2}\pi]$,
evaluate
$\cos^2A + (\cot B)(\sin A)$.
I haven't been able to figure out the angle for $B$.
Given $\tan A = \sqrt3$ and $\csc B = -\sqrt2$ where $A,B \in[\pi,\frac{3}{2}\pi]$,
evaluate
$\cos^2A + (\cot B)(\sin A)$.
I haven't been able to figure out the angle for $B$.
$\csc B = -\sqrt 2 \implies \frac 1 {\sin B}=-\sqrt 2\implies\frac {-1} {\sqrt 2}=\sin B$ Certainly now you can figure out what $B$ is.
Using the identity $1+\tan^2A=\sec^2A$, you have that $\sec^2A=4$. i.e $\cos^2A=\frac{1}{4}$. So $\cos A=-\frac{1}{2}$, since $A$ is in the third quadrant. From this we get $\sin A=-\frac{\sqrt3}{2}$, again since $A$ is in the third quadrant.
Now, using the identity $1+\cot^2B=\csc^2B$, you get $\cot B=1$. This is positive since $\tan$ is positive in the third quadrant (which means $\cot$ is also positive), and so the result follows.
Hints: We can find an explicit expression without knowing the angles (the angles are, however, not hard to find).
For example, recall the identity $1+\tan^2 A=\sec^2 A$. If you know $\tan A$, then you know $\sec^2 A$, and therefore you know $\cos^2 A$.
If you know $\cos^2 A$, you know $\sin^2 A$. Therefore you almost know $\sin A$, apart from sign. But since $A$ is in the third quadrant, you know the sign of $\sin A$.
You know $\csc B$, and want to know $\cot B$. There are various approaches. For example, from $\sin^2x+\cos^2 x=1$, dividing both sides by $\sin^2 x$, you get $1+\cot^2 x=\csc^2 x$. If you know $\csc B$, you can find $\cot^2 B$, and then, $\cot B$. Again, you have to worry about sign. But that's not hard, $\tan$ is positive in the third quadrant.
Or else you know $\csc B$, so you know $\sin^2 B$, so you know $\cos^2 B$, so you know $\cot^2 B$.
Comment: We know that $\csc B=-\sqrt{2}$. Thus $\sin B=-1/\sqrt{2}$, a familiar quantity. We know that the sine of $\pi/4$ is $1/\sqrt{2}$. So halfway between $\pi$ and $3\pi/2$, the sine is $-1/\sqrt{2}$.