Question

let θ be an angle in quadrant iii such that tan θ = 4/5. find the exact values of cos θ and csc θ. cos θ = csc θ =

Explanation:

Step1: Recall the relationship between tangent and sine - cosine

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Given $\tan\theta = \frac{4}{3}$ and $\theta$ is in quadrant III, where $\sin\theta<0$ and $\cos\theta<0$. Also, $\tan\theta=\frac{y}{x}=\frac{4}{3}$, so we can assume $y = - 4$ and $x=-3$ (negative because in quadrant III).

Step2: Find the value of the hypotenuse $r$

Using the Pythagorean theorem $r=\sqrt{x^{2}+y^{2}}$. Substitute $x = - 3$ and $y=-4$ into it, we get $r=\sqrt{(-3)^{2}+(-4)^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Calculate the cosine value

The formula for cosine is $\cos\theta=\frac{x}{r}$. Substituting $x=-3$ and $r = 5$, we have $\cos\theta=-\frac{3}{5}$.

Step4: Calculate the cosecant value

First, $\sin\theta=\frac{y}{r}=-\frac{4}{5}$. Since $\csc\theta=\frac{1}{\sin\theta}$, then $\csc\theta=-\frac{5}{4}$.

Answer:

$\cos\theta=-\frac{3}{5}$, $\csc\theta=-\frac{5}{4}$