If sin of theta equals 3/8 and theta is in quadrant II. what are cos,...

${\displaystyle \sin \theta =\frac{3}{8}}$
By rearranging the Pythagorean theorem, we can calculate the adjacent side, b, since we already have the side opposite to ${\displaystyle \theta }$, ${\displaystyle 3}$, and the hypotenuse, ${\displaystyle 8}$.
${\displaystyle a^2+b^2=c^2}$
${\displaystyle b^2=c^2-a^2}$
${\displaystyle b^2=8^2-3^2}$
${\displaystyle b^2=64-9}$
${\displaystyle b^2=55}$
${\displaystyle b=-\sqrt{55}}$
From this, we can determine that the side that is next to it has a length of ${\displaystyle -\sqrt{55}}$ units, since the x-axis is negative in the second quadrant. Therefore, we can infer that ${\displaystyle \cos \theta =-\frac{\sqrt{55}}{8}}$ and ${\displaystyle \tan \theta =-\frac{3}{\sqrt{55}}}$
Now, using the reciprocal identities, we must take the reciprocals of forces, sine, and cosine.
${\displaystyle \csc \theta =\frac{1}{\sin \theta }}$
${\displaystyle \sec \theta =\frac{1}{\cos \theta }}$
${\displaystyle \cot \theta =\frac{1}{\tan \theta }}$
So,
${\displaystyle \csc \theta =\frac{8}{3},\sec \theta =-\frac{8}{\sqrt{55}}\text{and}\cot \theta =-\frac{\sqrt{55}}{3}}$
In conclusion, the six trigonometric ratios are:
${\displaystyle \sin \theta =\frac{3}{8}}$
${\displaystyle \cos \theta =-\frac{\sqrt{55}}{8}}$
${\displaystyle \tan \theta =-\frac{3}{\sqrt{55}}}$
${\displaystyle \csc \theta =\frac{8}{3}}$
${\displaystyle \sec \theta =-\frac{8}{\sqrt{55}}}$
${\displaystyle \cot \theta =-\frac{\sqrt{55}}{3}}$

${\displaystyle \sin \theta =\frac{3}{8}\to \left(1\right)}$
${\displaystyle •\csc \theta =\frac{1}{\sin \theta }=\frac{8}{3}\to \left(2\right)}$
${\displaystyle •\cos \theta =\pm \sqrt{1-{\sin }^2\theta }}$
${\displaystyle \times \times x=-\sqrt{1-\frac{9}{64}}}$
${\displaystyle \times \times x=-\sqrt{\frac{55}{64}}}$
${\displaystyle \Rightarrow \cos \theta =-\frac{\sqrt{55}}{8}\to \left(3\right)}$
${\displaystyle •\sec \theta =\frac{1}{\cos \theta }=-\frac{8}{\sqrt{55}}\to \left(4\right)}$
${\displaystyle •\sec \theta =\frac{1}{\cos \theta }=-\frac{8}{\sqrt{55}}\to \left(4\right)}$
${\displaystyle •\tan \theta =\frac{\sin \theta }{\cos \theta }}$
${\displaystyle \times \times =\frac{3}{8}\times -\frac{8}{\sqrt{55}}=-\frac{3}{\sqrt{55}}\to \left(5\right)}$
${\displaystyle •\cot \theta =\frac{1}{\tan \theta }=-\frac{\sqrt{55}}{3}\to \left(6\right)}$