Question

use identities to find the exact value of the trigonometric expression below.$\frac{1}{sin^{2}\frac{pi}{16}} - \frac{1}{\tan^{2}\frac{pi}{16}}$$\frac{1}{sin^{2}\frac{pi}{16}} - \frac{1}{\tan^{2}\frac{pi}{16}} = square$(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Rewrite $\tan^2 x$ as $\frac{\sin^2 x}{\cos^2 x}$

Let $x = \frac{\pi}{16}$. The expression becomes:
$$\frac{1}{\sin^2 x} - \frac{1}{\frac{\sin^2 x}{\cos^2 x}}$$

Step2: Simplify the second fraction

$$\frac{1}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x}$$

Step3: Combine the fractions

$$\frac{1 - \cos^2 x}{\sin^2 x}$$

Step4: Use Pythagorean identity $1-\cos^2 x=\sin^2 x$

$$\frac{\sin^2 x}{\sin^2 x}$$

Step5: Simplify the expression

$$1$$

Answer:

$1$