5. 0/4 points details my notes larpcalclimtx3 4.3.046. use the given function value and the trigonometric…

Explanation:

Step1: Recall the reciprocal identity

We know that $\sec\beta=\frac{1}{\cos\beta}$. Given $\cos\beta = \frac{\sqrt{17}}{9}$, then $\sec\beta=\frac{9}{\sqrt{17}}=\frac{9\sqrt{17}}{17}$.

Step2: Use the Pythagorean identity

The Pythagorean identity is $\sin^{2}\beta+\cos^{2}\beta = 1$. So $\sin\beta=\sqrt{1 - \cos^{2}\beta}$. Substituting $\cos\beta=\frac{\sqrt{17}}{9}$, we have $\sin\beta=\sqrt{1-\left(\frac{\sqrt{17}}{9}\right)^{2}}=\sqrt{1-\frac{17}{81}}=\sqrt{\frac{81 - 17}{81}}=\sqrt{\frac{64}{81}}=\frac{8}{9}$.

Step3: Recall the tangent - cosine - sine relationship

The identity $\tan\beta=\frac{\sin\beta}{\cos\beta}$. Since $\sin\beta=\frac{8}{9}$ and $\cos\beta=\frac{\sqrt{17}}{9}$, then $\tan\beta=\frac{\frac{8}{9}}{\frac{\sqrt{17}}{9}}=\frac{8}{\sqrt{17}}=\frac{8\sqrt{17}}{17}$.

Step4: Use the co - function identity

The co - function identity $\sin(90^{\circ}-\beta)=\cos\beta$. Given $\cos\beta=\frac{\sqrt{17}}{9}$, so $\sin(90^{\circ}-\beta)=\frac{\sqrt{17}}{9}$.

Answer:

(a) $\frac{9\sqrt{17}}{17}$ (b) $\frac{8}{9}$ (c) $\frac{8\sqrt{17}}{17}$ (d) $\frac{\sqrt{17}}{9}$