Question

Can somebody please help me? (PreCalc)

Answer 1

$\csc\theta=\dfrac1{\sin\theta}=\dfrac{\sqrt{17}}}4\implies\sin\theta=\dfrac4{\sqrt{17}}$

Recall that $\sin^2\theta+\cos^2\theta=1$, which means

$\cos\theta=\pm\sqrt{1-\sin^2\theta}=\pm\sqrt{1-\dfrac{16}{17}}=\pm\dfrac1{\sqrt{17}}$

Since $\tan\theta=\dfrac{\sin\theta}{\cos\theta}>0$, and $\sin\theta>0$, it follows that $\cos\theta>0$ too, so you take the positive root:

$\cos\theta=\dfrac1{\sqrt{17}}$

Therefore

$\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac1{\sqrt{17}}}{\frac4{\sqrt{17}}}=\dfrac14$