answer:Given tan 2alpha = frac{cos alpha}{2-sin alpha}, we start by expressing tan 2alpha in terms of sin alpha and cos alpha: [ tan 2alpha = frac{sin 2alpha}{cos 2alpha} = frac{2sin alpha cos alpha}{1-2sin^2 alpha} ] Therefore, we have: [ frac{2sin alpha cos alpha}{1-2sin^2 alpha} = frac{cos alpha}{2-sin alpha} ] Given that alpha in (0, frac{pi}{2}), it implies cos alpha neq 0. Thus, we can safely divide both sides of the equation by cos alpha without worrying about division by zero. This simplifies our equation to: [ 2sin alpha (2-sin alpha) = 1-2sin^2 alpha ] Expanding and rearranging the terms gives us a quadratic equation in sin alpha: [ 4sin alpha - 2sin^2 alpha = 1 - 2sin^2 alpha ] Simplifying further, we find: [ 4sin alpha = 1 ] Thus, we have: [ sin alpha = frac{1}{4} ] Next, we find cos alpha using the Pythagorean identity: [ cos alpha = sqrt{1-sin^2 alpha} = sqrt{1-left(frac{1}{4}right)^2} = sqrt{1-frac{1}{16}} = sqrt{frac{15}{16}} = frac{sqrt{15}}{4} ] Finally, we calculate tan alpha: [ tan alpha = frac{sin alpha}{cos alpha} = frac{frac{1}{4}}{frac{sqrt{15}}{4}} = frac{1}{sqrt{15}} = frac{sqrt{15}}{15} ] Therefore, the correct answer is boxed{text{A: }frac{sqrt{15}}{15}}.

answer:1. **Identify the Solving Strategy**: We need to determine the area within the square ABCD where the angle angle BPC > 90^circ. This involves understanding the geometric properties of the semicircle omega with diameter overline{BC}. Thales' Theorem helps us here, stating that any angle subtended by the diameter of a semicircle is a right angle. 2. **Understanding Thales' Theorem**: According to Thales' Theorem: - angle BPC < 90^circ if point P lies inside omega (the semicircle). - angle BPC = 90^circ if point P lies on omega (the circumference of the semicircle). - angle BPC > 90^circ if point P lies outside omega. 3. **Calculation of the Area of Interest**: We compute the area within the square ABCD where angle BPC > 90^circ, which corresponds to the area outside omega. 4. **Calculate Area of omega**: The diameter overline{BC} of omega is equal to the side length of the square, which we set to 1 unit for simplicity. Thus, the radius r of omega is: [ r = frac{1}{2} ] The area of omega, which is a semicircle, is given by: [ text{Area of } omega = frac{1}{2} pi r^2 = frac{1}{2} pi left(frac{1}{2}right)^2 = frac{1}{2} pi cdot frac{1}{4} = frac{pi}{8} ] 5. **Calculate Total Area of Square**: The area of the square ABCD with side length 1 is: [ text{Area of } ABCD = 1 times 1 = 1 ] 6. **Compute the Probability**: Since the point P is selected uniformly at random inside the square, the probability is proportional to the area. Thus, the probability that P lies inside omega is: [ text{Probability}(P text{ inside } omega) = frac{text{Area of }omega}{text{Area of }ABCD} = frac{frac{pi}{8}}{1} = frac{pi}{8} ] 7. **Probability of angle BPC > 90^circ**: The probability that angle BPC > 90^circ is the complement of the probability that P lies inside omega: [ text{Probability}(angle BPC > 90^circ) = 1 - frac{pi}{8} ] **Conclusion**: Thus, the probability that angle BPC > 90^circ is: [ boxed{1 - frac{pi}{8}} ]

answer:This problem requires us to understand the new definition provided and use it to find the elements of the set A*B. The new definition is a type that has been frequently tested in recent exams. According to the definition, A*B={z|z=xy, xin A, yin B}. Given that A={1,2} and B={0,2}, we can find the elements of A*B as follows: - For x=1 and y=0, we have z=1*0=0. - For x=1 and y=2, we have z=1*2=2. - For x=2 and y=0, we have z=2*0=0. However, since sets do not contain duplicate elements, we discard this result. - For x=2 and y=2, we have z=2*2=4. Therefore, the set A*B contains the elements {0, 2, 4}. The sum of all elements in the set A*B is 0+2+4=boxed{6}.

answer:To solve this, we break the integral into two parts: ( int_{0}^{1} (2x+ sqrt{1-x^{2}})dx = int_{0}^{1} 2xdx + int_{0}^{1} sqrt{1-x^{2}}dx ). By the geometric interpretation of the definite integral, ( int_{0}^{1} sqrt{1-x^{2}}dx ) represents a quarter of the area of a unit circle, which is ( frac{pi}{4} ). For the first part, ( int_{0}^{1} 2xdx = x^{2} |_{0}^{1} = 1 ). Therefore, ( int_{0}^{1} (2x+ sqrt{1-x^{2}})dx = 1+ frac{pi}{4} ). Hence, the answer is ( boxed{1+ frac{pi}{4}} ). This problem tests the operational properties of definite integrals and their geometric meanings, focusing on computational skills. It is considered a basic question.