Find the exact value of each of the remaining trigonometric functions of $\theta$ . $\tan \theta = -\frac{1}{5},\quad \sec \theta < 0$

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Algebra 2

Simplify expressions using trigonometric identities

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Q. Find the exact value of each of the remaining trigonometric functions of

\theta

\tan \theta = -\frac{1}{5},\quad \sec \theta < 0

Given Information: We are given that $\tan(\theta) = -\frac{1}{5}$ . Since $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ , we can represent $\sin(\theta)$ and $\cos(\theta)$ as two sides of a right triangle where $\sin(\theta)$ is the opposite side and $\cos(\theta)$ is the adjacent side to the angle $\theta$ . We can use the Pythagorean theorem to find the hypotenuse. Let's assume the opposite side ( $\sin(\theta)$ ) is $-1$ (since $\tan$ is negative, and we are in a quadrant where $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ $0$ is also negative) and the adjacent side ( $\cos(\theta)$ ) is $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ $2$ . The hypotenuse will be the square root of the sum of the squares of the opposite and adjacent sides.
Pythagorean Theorem Calculation: Using the Pythagorean theorem, we calculate the hypotenuse as $\sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$ . This is the value of the hypotenuse, which corresponds to the value of $r$ (the radius in trigonometric functions).
Finding $\cos(\theta)$ and $\sin(\theta)$ : Now we can find the values of $\sin(\theta)$ and $\cos(\theta)$ . Since we are in a quadrant where $\sec(\theta)$ is negative, this means $\cos(\theta)$ must be negative (because $\sec(\theta)$ is the reciprocal of $\cos(\theta)$ ). Therefore, $\cos(\theta) = -\frac{5}{\sqrt{26}}$ . To rationalize the denominator, we multiply the numerator and the denominator by $\sqrt{26}$ to get $\sin(\theta)$ $0$ .
Calculating cot(theta): To find $\sin(\theta)$ , we use the opposite side over the hypotenuse. Since the opposite side is $-1$ and the hypotenuse is $\sqrt{26}$ , $\sin(\theta) = -1/\sqrt{26}$ . Rationalizing the denominator, we get $\sin(\theta) = -\sqrt{26}/26$ .
Determining sec(theta): We already have $\tan(\theta) = -\frac{1}{5}$ . Now we need to find $\cot(\theta)$ , which is the reciprocal of $\tan(\theta)$ . Therefore, $\cot(\theta) = -5$ .
Calculating $\csc(\theta)$ : To find $\sec(\theta)$ , which is the reciprocal of $\cos(\theta)$ , we take the reciprocal of $-\frac{5\sqrt{26}}{26}$ to get $\sec(\theta) = -\frac{26}{5\sqrt{26}}$ . Rationalizing the denominator, we get $\sec(\theta) = -\frac{26\sqrt{26}}{130} = -\frac{2\sqrt{26}}{10}$ .
Calculating csc(theta): To find sec(theta), which is the reciprocal of cos(theta), we take the reciprocal of $-5\sqrt{26}/26$ to get sec(theta) = $-26/(5\sqrt{26})$ . Rationalizing the denominator, we get sec(theta) = $-26\sqrt{26}/130 = -2\sqrt{26}/10$ .Finally, to find csc(theta), which is the reciprocal of sin(theta), we take the reciprocal of $-\sqrt{26}/26$ to get csc(theta) = $-26/\sqrt{26}$ . Rationalizing the denominator, we get csc(theta) = $-26\sqrt{26}/26 = -\sqrt{26}$ .