Answer
$b=\sqrt {c^{2}-a^{2}}=\sqrt {2^{2}-1^{2}}=\sqrt {3}$
$\sin B=\frac{\sqrt {3}}{2}$
$\cos B=\frac{1}{2}$
$\tan B=\sqrt 3$
$\csc B=\frac{2\sqrt {3}}{3}$
$\sec B=2$
$\cot B=\frac{\sqrt {3}}{3}$
Work Step by Step
$b=\sqrt {c^{2}-a^{2}}=\sqrt {2^{2}-1^{2}}=\sqrt {3}$
$\sin B=\frac{\text{side opposite}}{\text{hypotenuse}}=\frac{b}{c}=\frac{\sqrt {3}}{2}$
$\cos B=\frac{\text{side adjacent}}{\text{hypotenuse}}=\frac{a}{c}=\frac{1}{2}$
$\tan B=\frac{\text{side opposite}}{\text{side adjacent}}=\frac{b}{a}=\frac{\sqrt {3}}{1}=\sqrt 3$
The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so we have
$\csc B=\frac{2}{\sqrt {3}}=\frac{2\sqrt {3}}{3}$
$\sec B=\frac{2}{1}=2$
$\cot B=\frac{1}{\sqrt {3}}=\frac{\sqrt {3}}{3}$
