Question

1. given the ratio for tan θ, find the remaining ratios.

$4^2 + 5^2 = x^2$
$16 + 25 = x^2$
$sqrt{41} = sqrt{x^2}$
$x = sqrt{41}$
$\sin \theta = $ $\csc \theta = $
$\cos \theta = $ $\sec \theta = $
$\tan \theta = \frac{4}{5}$ $\cot \theta = $
give the exact ratios for each functio

Explanation:

Step1: Identify sides relative to $\theta$

For angle $\theta$: opposite = 4, adjacent = 5, hypotenuse = $\sqrt{41}$

Step2: Calculate $\sin\theta$

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{\sqrt{41}} = \frac{4\sqrt{41}}{41}$

Step3: Calculate $\csc\theta$ (reciprocal of $\sin\theta$)

$\csc\theta = \frac{1}{\sin\theta} = \frac{\sqrt{41}}{4}$

Step4: Calculate $\cos\theta$

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{\sqrt{41}} = \frac{5\sqrt{41}}{41}$

Step5: Calculate $\sec\theta$ (reciprocal of $\cos\theta$)

$\sec\theta = \frac{1}{\cos\theta} = \frac{\sqrt{41}}{5}$

Step6: Calculate $\cot\theta$ (reciprocal of $\tan\theta$)

$\cot\theta = \frac{1}{\tan\theta} = \frac{5}{4}$

Answer:

$\sin\theta = \frac{4\sqrt{41}}{41}$
$\csc\theta = \frac{\sqrt{41}}{4}$
$\cos\theta = \frac{5\sqrt{41}}{41}$
$\sec\theta = \frac{\sqrt{41}}{5}$
$\cot\theta = \frac{5}{4}$