Question

1. find the values of the six trigonometric functions for $\theta$.

(right triangle with legs 12, $x$, hypotenuse 16, angle $\theta$ at the leg 12 vertex)
$x^2 + 12^2 = 16^2$
$x^2 + 144 = 256$
$x^2 = 112$
$x = 4\sqrt{7}$
$\sin \theta = \frac{4\sqrt{7}}{16}$
$\csc\theta = $
$\cos \theta = \frac{12}{16} = \frac{3}{4}$
$\sec\theta = \frac{4}{3}$
$\tan \theta = \frac{4\sqrt{7}}{12}$
$\cot \theta = $

Explanation:

Step1: Define cscθ as reciprocal of sinθ

$\csc\theta = \frac{1}{\sin\theta}$

Step2: Substitute $\sin\theta = \frac{\sqrt{7}}{4}$

$\csc\theta = \frac{1}{\frac{\sqrt{7}}{4}} = \frac{4}{\sqrt{7}} = \frac{4\sqrt{7}}{7}$

Step3: Simplify tanθ first

$\tan\theta = \frac{4\sqrt{7}}{12} = \frac{\sqrt{7}}{3}$

Step4: Define cotθ as reciprocal of tanθ

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

Step5: Substitute $\tan\theta = \frac{\sqrt{7}}{3}$

$\cot\theta = \frac{1}{\frac{\sqrt{7}}{3}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

Answer:

$\csc\theta = \frac{4\sqrt{7}}{7}$
$\cot\theta = \frac{3\sqrt{7}}{7}$