Question

Please help me thank you

Answer 1

Answer:

$\large\boxed{-\dfrac{8}{\sqrt{63}}}$

Step-by-step explanation:

$\csc\theta=\dfrac{1}{\sin\theta}\\\\\cos\theta=-\dfrac{1}{8}\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta+\left(-\dfrac{1}{8}\right)^2=1\\\\\sin^2\theta+\dfrac{1}{64}=1\qquad\text{subtract}\ \dfrac{1}{64}\ \text{from both sides}\\\\\sin^2\theta=\dfrac{64}{64}-\dfrac{1}{64}\\\\\sin^2\theta=\dfrac{63}{64}\to\sin\theta=\pm\sqrt{\dfrac{63}{64}}\\\\180^o$

$\csc\theta=\dfrac{1}{-\frac{\sqrt{63}}{8}}=-\dfrac{8}{\sqrt{63}}$