let θ be an angle in quadrant ii such that tanθ = - 5/3. find the exact values of cosθ and cscθ. cosθ = cscθ =

Explanation:

Step1: Recall the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Given $\tan\theta =-\frac{5}{3}=\frac{\sin\theta}{\cos\theta}$, so $\sin\theta =-\frac{5}{3}\cos\theta$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

Substitute $\sin\theta =-\frac{5}{3}\cos\theta$ into $\sin^{2}\theta+\cos^{2}\theta = 1$. We get $\left(-\frac{5}{3}\cos\theta\right)^{2}+\cos^{2}\theta=1$. Expanding gives $\frac{25}{9}\cos^{2}\theta+\cos^{2}\theta = 1$. Combining like - terms: $\left(\frac{25 + 9}{9}\right)\cos^{2}\theta=1$, so $\frac{34}{9}\cos^{2}\theta = 1$, and $\cos^{2}\theta=\frac{9}{34}$. Since $\theta$ is in quadrant II, $\cos\theta<0$. Thus, $\cos\theta=-\frac{3}{\sqrt{34}}=-\frac{3\sqrt{34}}{34}$.

Step3: Find $\sin\theta$

Since $\sin\theta =-\frac{5}{3}\cos\theta$, substituting $\cos\theta =-\frac{3}{\sqrt{34}}$ gives $\sin\theta=\frac{5}{\sqrt{34}}=\frac{5\sqrt{34}}{34}$.

Step4: Recall the definition of $\csc\theta$

$\csc\theta=\frac{1}{\sin\theta}$. Since $\sin\theta=\frac{5\sqrt{34}}{34}$, then $\csc\theta=\frac{\sqrt{34}}{5}$.

Answer:

$\cos\theta=-\frac{3\sqrt{34}}{34}$ $\csc\theta=\frac{\sqrt{34}}{5}$