Question

Smart people. Please click here. I attached a pic of the problem

Answer 1

Look at the picture.

$r=\sqrt{x^2+y^2}$

$\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}\\\\\sec\theta=\dfrac{r}{x}\\\\\csc\thera=\dfrac{r}{y}$

We have the point $\left(\dfrac{8}{17},\ \dfrac{15}{17}\right)\to x=\dfrac{8}{17},\ y=\dfrac{15}{17}$.

Calculate r:

$r=\sqrt{\left(\dfrac{8}{17}\right)^2+\left(\dfrac{15}{17}\right)^2}=\sqrt{\dfrac{64}{289}+\dfrac{225}{289}}=\sqrt{\dfrac{289}{289}}=\sqrt1=1$

$\sin\theta=\dfrac{\frac{15}{17}}{1}=\dfrac{15}{17}\\\\\cos\theta=\dfrac{\frac{8}{17}}{1}=\dfrac{8}{17}\\\\\tan\theta=\dfrac{\frac{15}{17}}{\frac{8}{17}}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{17}\cdot\dfrac{17}{8}=\dfrac{15}{8}\\\\\cot\theta=\dfrac{\frac{8}{17}}{\frac{15}{17}}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{17}\cdot\dfrac{17}{15}=\dfrac{8}{15}\\\\\sec\theta=\dfrac{1}{\frac{8}{17}}=1:\dfrac{8}{17}=\dfrac{17}{8}\\\\\csc\theta=\dfrac{1}{\frac{15}{17}}=1:\dfrac{15}{17}=\dfrac{17}{15}$