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

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Q. Find the exact value of each of the remaining trigonometric functions of θ\theta.tan⁡θ=−15,sec⁡θ<0\tan \theta = -\frac{1}{5},\quad \sec \theta < 0
  1. Given Information: We are given that tan⁡(θ)=−15\tan(\theta) = -\frac{1}{5}. Since tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, we can represent sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) as two sides of a right triangle where sin⁡(θ)\sin(\theta) is the opposite side and cos⁡(θ)\cos(\theta) is the adjacent side to the angle θ\theta. We can use the Pythagorean theorem to find the hypotenuse. Let's assume the opposite side (sin⁡(θ)\sin(\theta)) is −1-1 (since tan⁡\tan is negative, and we are in a quadrant where tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}00 is also negative) and the adjacent side (cos⁡(θ)\cos(\theta)) is tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}22. The hypotenuse will be the square root of the sum of the squares of the opposite and adjacent sides.
  2. Pythagorean Theorem Calculation: Using the Pythagorean theorem, we calculate the hypotenuse as (−1)2+52=1+25=26\sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}. This is the value of the hypotenuse, which corresponds to the value of rr (the radius in trigonometric functions).
  3. Finding cos⁡(θ)\cos(\theta) and sin⁡(θ)\sin(\theta): Now we can find the values of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta). Since we are in a quadrant where sec⁡(θ)\sec(\theta) is negative, this means cos⁡(θ)\cos(\theta) must be negative (because sec⁡(θ)\sec(\theta) is the reciprocal of cos⁡(θ)\cos(\theta)). Therefore, cos⁡(θ)=−526\cos(\theta) = -\frac{5}{\sqrt{26}}. To rationalize the denominator, we multiply the numerator and the denominator by 26\sqrt{26} to get sin⁡(θ)\sin(\theta)00.
  4. Calculating cot(theta): To find sin⁡(θ)\sin(\theta), we use the opposite side over the hypotenuse. Since the opposite side is −1-1 and the hypotenuse is 26\sqrt{26}, sin⁡(θ)=−1/26\sin(\theta) = -1/\sqrt{26}. Rationalizing the denominator, we get sin⁡(θ)=−26/26\sin(\theta) = -\sqrt{26}/26.
  5. Determining sec(theta): We already have tan⁡(θ)=−15\tan(\theta) = -\frac{1}{5}. Now we need to find cot⁡(θ)\cot(\theta), which is the reciprocal of tan⁡(θ)\tan(\theta). Therefore, cot⁡(θ)=−5\cot(\theta) = -5.
  6. Calculating csc⁡(θ)\csc(\theta): To find sec⁡(θ)\sec(\theta), which is the reciprocal of cos⁡(θ)\cos(\theta), we take the reciprocal of −52626-\frac{5\sqrt{26}}{26} to get sec⁡(θ)=−26526\sec(\theta) = -\frac{26}{5\sqrt{26}}. Rationalizing the denominator, we get sec⁡(θ)=−2626130=−22610\sec(\theta) = -\frac{26\sqrt{26}}{130} = -\frac{2\sqrt{26}}{10}.
  7. Calculating csc(theta): To find sec(theta), which is the reciprocal of cos(theta), we take the reciprocal of −526/26-5\sqrt{26}/26 to get sec(theta) = −26/(526)-26/(5\sqrt{26}). Rationalizing the denominator, we get sec(theta) = −2626/130=−226/10-26\sqrt{26}/130 = -2\sqrt{26}/10.Finally, to find csc(theta), which is the reciprocal of sin(theta), we take the reciprocal of −26/26-\sqrt{26}/26 to get csc(theta) = −26/26-26/\sqrt{26}. Rationalizing the denominator, we get csc(theta) = −2626/26=−26-26\sqrt{26}/26 = -\sqrt{26}.

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