Questions: Refer to the following figures. In each case, express the indicated trigonometric values as functions of (x). Rationalize any denominators containing radicals. (a) (csc alpha, sec alpha, cot alpha) (csc alpha=) (sec alpha=) (cot alpha=) (b) (sin ^2 alpha+cos ^2 alpha+tan ^2 alpha)

Refer to the following figures. In each case, express the indicated trigonometric values as functions of (x). Rationalize any denominators containing radicals.
(a) (csc alpha, sec alpha, cot alpha)

(csc alpha=)

(sec alpha=)

(cot alpha=)

(b) (sin ^2 alpha+cos ^2 alpha+tan ^2 alpha)

Transcript text: Refer to the following figures. In each case, express the indicated trigonometric values as functions of $x$. Rationalize any denominators containing radicals. (a) $\csc \alpha, \sec \alpha, \cot \alpha$ \[ \begin{array}{l} \csc \alpha=\square \\ \sec \alpha=\square \\ \cot \alpha=\square \end{array} \] (b) $\sin ^{2} \alpha+\cos ^{2} \alpha+\tan ^{2} \alpha$ $\square$

Solution

Solution Steps

Step 1: Finding the hypotenuse of the first triangle

We have a right triangle with legs of length 3 and 2x. Let h be the length of the hypotenuse. By the Pythagorean theorem, we have $h^2 = 3^2 + (2x)^2 = 9 + 4x^2$, so $h = \sqrt{9+4x^2}$.

Step 2: Calculating csc α, sec α, and cot α

$\csc \alpha = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{\sqrt{9+4x^2}}{3}$
$\sec \alpha = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{9+4x^2}}{2x}$
$\cot \alpha = \frac{\text{adjacent}}{\text{opposite}} = \frac{2x}{3}$

Step 3: Calculating sin² α + cos² α + tan² α

We know the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$. Therefore, we only need to find $\tan^2 \alpha$. $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{2x}$. Thus, $\tan^2 \alpha = \frac{9}{4x^2}$.

Then, $\sin^2 \alpha + \cos^2 \alpha + \tan^2 \alpha = 1 + \frac{9}{4x^2} = \frac{4x^2 + 9}{4x^2}$.