answer:Solution: (1) Since sin theta + cos theta = frac{1}{5}, and theta in (0,pi), squaring both sides gives 1 + 2sin theta cos theta = frac{1}{25}, thus sin theta cos theta = -frac{12}{25}. From these, we find sin theta = frac{4}{5} and cos theta = -frac{3}{5}, therefore tan theta = frac{sin theta}{cos theta} = -frac{4}{3}. So, boxed{tan theta = -frac{4}{3}}. (2) frac{1-2sin theta cos theta}{cos^2 theta - sin^2 theta} = frac{(cos theta - sin theta)^2}{(cos theta + sin theta) cdot (cos theta - sin theta)} = frac{cos theta - sin theta}{cos theta + sin theta} = frac{1 - tan theta}{1 + tan theta} = -7. Thus, boxed{-7}.

answer:1. Let the vertices of the triangle be ( P ), ( Q ), and ( R ). We are given that ( PQ = 6 ) and the radius of the circumcircle ( R ) is ( 5 ). 2. Let (angle PRQ) be denoted by (gamma). 3. By the Law of Sines, the sine of angle (gamma) can be calculated using: [ sin gamma = frac{PQ}{2R} ] Substituting the given values: [ sin gamma = frac{6}{2 times 5} = frac{6}{10} = frac{3}{5} ] 4. Now, we calculate (cos gamma). The value of (cos gamma) can be determined using the Pythagorean identity for trigonometric functions: [ cos^2 gamma + sin^2 gamma = 1 cos^2 gamma + left(frac{3}{5}right)^2 = 1 cos^2 gamma + frac{9}{25} = 1 cos^2 gamma = 1 - frac{9}{25} cos^2 gamma = frac{25}{25} - frac{9}{25} cos^2 gamma = frac{16}{25} cos gamma = pm frac{4}{5} ] Since (gamma) could be either acute or obtuse, (cos gamma) has two possible values: (frac{4}{5}) or (-frac{4}{5}). 5. We need to find the length of the chord ( P'Q' ). The chord ( P'Q' ) connects points on the circumcircle that subtend the same angle (gamma) at the center of the circle. Using the fact that a chord of a circle subtends an angle of (pi - gamma) at the circumference, we have: [ text{Length of chord } P'Q' = 2R sin left( frac{pi - gamma}{2} right) ] Simplifying further: [ sin left( frac{pi - gamma}{2} right) = cos left( frac{gamma}{2} right) ] Therefore: [ P'Q' = 2R cos left( frac{gamma}{2} right) ] 6. We use the half-angle formula for cosine to find ( cos left( frac{gamma}{2} right) ): [ cos left( frac{gamma}{2} right) = sqrt{frac{1 + cos gamma}{2}} ] There are two cases to consider: **Case 1:** (cos gamma = frac{4}{5}): [ cos left( frac{gamma}{2} right) = sqrt{frac{1 + frac{4}{5}}{2}} = sqrt{frac{frac{9}{5}}{2}} = sqrt{frac{9}{10}} = frac{3}{sqrt{10}} ] Thus: [ P'Q' = 2 times 5 times frac{3}{sqrt{10}} = 10 times frac{3}{sqrt{10}} = 3 sqrt{10} ] **Case 2:** (cos gamma = -frac{4}{5}): [ cos left( frac{gamma}{2} right) = sqrt{frac{1 + left( -frac{4}{5} right)}{2}} = sqrt{frac{frac{1}{5}}{2}} = sqrt{frac{1}{10}} = frac{1}{sqrt{10}} ] Thus: [ P'Q' = 2 times 5 times frac{1}{sqrt{10}} = 10 times frac{1}{sqrt{10}} = sqrt{10} ] 7. Therefore, the length of the chord ( P'Q' ) can be either ( 3sqrt{10} ) or (sqrt{10} ). # Conclusion [ boxed{P'Q' = 3sqrt{10} text{ or } sqrt{10}} ]

answer:1. Calculate the volume of the rectangular prism: [ Volume = 12 times 4 times 18 = 864 text{ cubic inches} ] 2. Find the edge length of the cube that has the same volume: [ text{Edge length of the cube} = sqrt[3]{864} = 12 text{ inches} ] 3. Calculate the surface area of the cube: [ text{Surface area} = 6 times (text{Edge length})^2 = 6 times 12^2 = 6 times 144 = 864 text{ square inches} ] Conclusion: The surface area of the cube is boxed{864} square inches.

answer:To determine the number of ways to arrange the letters of the word TARTAR, we follow these steps: 1. **Count the total arrangements if all letters were unique:** Since there are 6 letters, the total arrangements would be 6!. 2. **Account for the non-uniqueness of the letters T, A, and R:** Each of these letters appears twice, so we must divide by 2! for each set of duplicate letters to correct for overcounting. This gives us a division by 2! times 2! times 2!. 3. **Calculate the final answer:** Combining these steps, we find the total number of unique arrangements as follows: [ dfrac{6!}{2! times 2! times 2!} = dfrac{720}{2 times 2 times 2} = dfrac{720}{8} = 90 ] Therefore, the number of ways to arrange the letters of the word TARTAR is boxed{90}.