💡Okay, so I have this problem here about a function f(x) that's defined on the positive real numbers, and it's monotonic. That means it's either always increasing or always decreasing, right? And for every x in (0, +∞), it satisfies this equation: f(f(x) - ln x) = e + 1. I need to find the interval where the real solution to the equation f(x) - f'(x) = e lies. The options are A: (0, 1/e), B: (1/e, 1), C: (1, e), D: (e, 4). Alright, let me try to break this down step by step. First, the function f is monotonic, so it's either strictly increasing or strictly decreasing. That might help later on when we're trying to figure out properties of f or its inverse.The key equation given is f(f(x) - ln x) = e + 1. Hmm, that looks a bit complicated. Maybe I can make a substitution to simplify it. Let me let t = f(x) - ln x. Then the equation becomes f(t) = e + 1. So, for every x, f(t) is a constant, e + 1. That suggests that t is a constant because if f(t) is the same for all x, then t must be the same for all x, right? Wait, no, that might not necessarily be true because f could be a constant function, but since f is monotonic, it can't be constant unless it's a constant function, which would only be possible if it's both non-increasing and non-decreasing, but monotonic usually implies strictly increasing or decreasing. So, f can't be constant. Therefore, t must be a constant. So, t is a constant value such that f(t) = e + 1.So, if t is a constant, then f(x) - ln x = t for all x. That means f(x) = ln x + t. So, f(x) is the natural logarithm of x plus some constant t. Now, since f is monotonic, and ln x is itself a monotonic function (it's increasing), then f(x) is also increasing because adding a constant doesn't change the monotonicity. So, that's consistent.Now, since t is a constant, and f(t) = e + 1, let's plug t into f(x). So, f(t) = ln t + t = e + 1. So, we have the equation ln t + t = e + 1. Hmm, I need to solve for t here. Let me write that down:ln t + t = e + 1.I need to find t such that this equation holds. Let me think about how to solve this. It's a transcendental equation, meaning it can't be solved algebraically, so I might need to use numerical methods or make an educated guess.Let me try plugging in t = e. Then, ln e = 1, so 1 + e = e + 1. Hey, that works! So, t = e. Therefore, f(x) = ln x + e.Great, so now I have the explicit form of f(x). Let me write that down:f(x) = ln x + e.Now, the problem asks for the interval where the real solution to the equation f(x) - f'(x) = e lies. So, let's compute f'(x). Since f(x) = ln x + e, the derivative f'(x) is 1/x. So, f'(x) = 1/x.Now, let's plug f(x) and f'(x) into the equation f(x) - f'(x) = e:ln x + e - (1/x) = e.Simplify this equation:ln x + e - 1/x = e.Subtract e from both sides:ln x - 1/x = 0.So, the equation reduces to:ln x - 1/x = 0.Let me denote this as a function h(x):h(x) = ln x - 1/x.We need to find the value of x where h(x) = 0. That is, find x such that ln x = 1/x.Now, to find the interval where this solution lies, I can evaluate h(x) at the endpoints of the given intervals and see where it crosses zero.Let's compute h(x) at some key points:1. At x = 1: h(1) = ln 1 - 1/1 = 0 - 1 = -1.2. At x = e: h(e) = ln e - 1/e = 1 - 1/e ≈ 1 - 0.3679 ≈ 0.6321.3. At x = 1/e: h(1/e) = ln(1/e) - 1/(1/e) = -1 - e ≈ -1 - 2.718 ≈ -3.718.4. At x = 4: h(4) = ln 4 - 1/4 ≈ 1.386 - 0.25 ≈ 1.136.So, let's see:- At x = 1, h(x) is -1.- At x = e, h(x) is approximately 0.6321.- At x = 1/e, h(x) is approximately -3.718.- At x = 4, h(x) is approximately 1.136.We are looking for where h(x) = 0. So, let's see where the function crosses zero.Between x = 1 and x = e:- At x = 1, h(x) = -1.- At x = e, h(x) ≈ 0.6321.Since h(x) is continuous (as both ln x and 1/x are continuous on (0, ∞)), and it goes from negative to positive between x = 1 and x = e, by the Intermediate Value Theorem, there must be some c in (1, e) where h(c) = 0.Similarly, let's check the other intervals:- Between x = 1/e and x = 1: At x = 1/e, h(x) ≈ -3.718. At x = 1, h(x) = -1. So, h(x) remains negative here, so no zero crossing.- Between x = e and x = 4: At x = e, h(x) ≈ 0.6321. At x = 4, h(x) ≈ 1.136. So, h(x) remains positive here, so no zero crossing.Therefore, the only interval where h(x) crosses zero is between x = 1 and x = e. So, the solution lies in (1, e).Hence, the correct answer is option C: (1, e).**Final Answer**boxed{C}

💡Alright, so I have this geometry problem here. It's about a right triangle ABC with the hypotenuse AB. On this hypotenuse, there's a square constructed externally, named ABPQ. Then, they define three angles: α is angle ACQ, β is angle QCP, and γ is angle PCB. The goal is to prove that cos β equals cos α times cos γ. Hmm, okay, let me try to visualize this.First, I need to sketch this out. I imagine triangle ABC, right-angled at C. So, AB is the hypotenuse. Then, on AB, we construct a square outwardly, so ABPQ is a square. That means all sides AB, BP, PQ, and QA are equal, and all angles are 90 degrees. So, points P and Q are outside the triangle ABC.Now, let's mark the points. Starting from A, moving to B, then constructing the square outward. So, from B, we go to P, then to Q, then back to A. So, point P is adjacent to B, and point Q is adjacent to A. Now, angles α, β, and γ are defined as follows: α is angle ACQ, which is the angle at point C between points A and Q. β is angle QCP, which is the angle at point C between points Q and P. γ is angle PCB, which is the angle at point C between points P and B.So, all three angles α, β, and γ are angles at point C, but between different points. So, starting from A, going to Q, then to P, then to B. So, the angles are kind of adjacent to each other at point C.I think I need to find some relationships between these angles. Maybe using trigonometric identities or properties of squares and right triangles. Since it's a square, all sides are equal, and all angles are right angles. So, perhaps there are some similar triangles or congruent triangles involved here.Let me consider triangle ACQ. Since Q is a corner of the square constructed on AB, AQ is equal to AB, and angle QAB is 90 degrees because it's a square. So, triangle ACQ has sides AC, CQ, and AQ. Similarly, triangle QCP has sides QC, CP, and PQ, and triangle PCB has sides PC, CB, and PB.Wait, maybe I can use the Law of Sines or Cosines in these triangles. Let me think about triangle ACQ first. If I can find expressions for the sides or angles, maybe I can relate them to the other triangles.Alternatively, maybe coordinate geometry can help. If I assign coordinates to the points, I can calculate the vectors and then find the angles using dot products or something like that. Let me try that approach.Let's place point C at the origin (0,0). Let me denote point A as (a,0) and point B as (0,b), so that triangle ABC is right-angled at C. Then, the hypotenuse AB goes from (a,0) to (0,b). Now, constructing square ABPQ externally on AB. So, starting from A, moving to B, then constructing the square outward.To find the coordinates of points P and Q, I need to figure out how the square is constructed. Since AB is from (a,0) to (0,b), the vector AB is (-a, b). To construct the square outward, we need to find points P and Q such that ABPQ is a square.The direction from B to P should be a 90-degree rotation of AB. Since AB is (-a, b), rotating this vector 90 degrees counterclockwise would give (-b, -a). So, point P would be at B plus this vector: (0,b) + (-b, -a) = (-b, b - a). Similarly, point Q would be at A plus the same rotated vector: (a,0) + (-b, -a) = (a - b, -a).Wait, let me confirm that. Rotating vector AB 90 degrees counterclockwise would change (x,y) to (-y,x). So, AB is (-a, b), so rotated 90 degrees counterclockwise would be (-b, -a). So, adding that to point B (0,b) gives (-b, b - a). Similarly, adding that to point A (a,0) gives (a - b, -a). So, points P(-b, b - a) and Q(a - b, -a).Now, with coordinates assigned, I can find the coordinates of points C, Q, P, and B. Point C is (0,0), Q is (a - b, -a), P is (-b, b - a), and B is (0,b).Now, let's find the vectors for the angles α, β, and γ.Angle α is angle ACQ, which is the angle at C between points A and Q. So, vectors CA and CQ. Vector CA is from C to A: (a,0). Vector CQ is from C to Q: (a - b, -a).Similarly, angle β is angle QCP, which is the angle at C between points Q and P. So, vectors CQ and CP. Vector CQ is (a - b, -a). Vector CP is from C to P: (-b, b - a).Angle γ is angle PCB, which is the angle at C between points P and B. So, vectors CP and CB. Vector CP is (-b, b - a). Vector CB is from C to B: (0,b).Now, to find the cosines of these angles, I can use the dot product formula. For two vectors u and v, the cosine of the angle between them is (u · v)/(|u||v|).Let me compute cos α first. Vectors CA = (a,0) and CQ = (a - b, -a).Dot product: (a)(a - b) + (0)(-a) = a(a - b).Magnitude of CA: sqrt(a² + 0²) = a.Magnitude of CQ: sqrt((a - b)² + (-a)²) = sqrt((a - b)² + a²).So, cos α = [a(a - b)] / [a * sqrt((a - b)² + a²)] = (a - b)/sqrt((a - b)² + a²).Simplify denominator: sqrt(a² - 2ab + b² + a²) = sqrt(2a² - 2ab + b²).So, cos α = (a - b)/sqrt(2a² - 2ab + b²).Okay, that's cos α.Now, let's compute cos γ. Vectors CP = (-b, b - a) and CB = (0,b).Dot product: (-b)(0) + (b - a)(b) = 0 + b(b - a) = b² - ab.Magnitude of CP: sqrt((-b)² + (b - a)²) = sqrt(b² + (b - a)²) = sqrt(b² + b² - 2ab + a²) = sqrt(2b² - 2ab + a²).Magnitude of CB: sqrt(0² + b²) = b.So, cos γ = (b² - ab)/[sqrt(2b² - 2ab + a²) * b] = (b - a)/sqrt(2b² - 2ab + a²).Wait, that's similar to cos α but with a and b swapped. Interesting.Now, let's compute cos β. Angle β is angle QCP, which is the angle between vectors CQ and CP.Vectors CQ = (a - b, -a) and CP = (-b, b - a).Dot product: (a - b)(-b) + (-a)(b - a) = -b(a - b) -a(b - a) = -ab + b² - ab + a² = a² - 2ab + b².Magnitude of CQ: sqrt((a - b)² + (-a)²) = sqrt(a² - 2ab + b² + a²) = sqrt(2a² - 2ab + b²).Magnitude of CP: sqrt((-b)² + (b - a)²) = sqrt(b² + b² - 2ab + a²) = sqrt(2b² - 2ab + a²).So, cos β = (a² - 2ab + b²)/[sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].Notice that the numerator is (a - b)², and the denominator is sqrt(2a² - 2ab + b²) times sqrt(2b² - 2ab + a²).Wait, let me see if I can factor the denominator. Let me denote D1 = sqrt(2a² - 2ab + b²) and D2 = sqrt(2b² - 2ab + a²). So, D1 = sqrt(a² + (a - b)²) and D2 = sqrt(b² + (b - a)²). Hmm, not sure if that helps.But let's look back at cos α and cos γ. We have:cos α = (a - b)/sqrt(2a² - 2ab + b²)cos γ = (b - a)/sqrt(2b² - 2ab + a²) = -(a - b)/sqrt(2b² - 2ab + a²)So, if I multiply cos α and cos γ, I get:cos α * cos γ = [(a - b)/sqrt(2a² - 2ab + b²)] * [-(a - b)/sqrt(2b² - 2ab + a²)] = -[(a - b)²]/[sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)]But from earlier, cos β is (a² - 2ab + b²)/[sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].Notice that (a² - 2ab + b²) is equal to (a - b)². So, cos β = (a - b)² / [sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].Comparing this with cos α * cos γ, which is -[(a - b)²]/[sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].Wait, so cos β is equal to negative of cos α * cos γ. But that contradicts the given statement that cos β = cos α * cos γ. Hmm, did I make a mistake somewhere?Let me check the signs. When I computed cos γ, I had vectors CP = (-b, b - a) and CB = (0,b). The dot product was (b - a)b = b² - ab. The magnitude of CP was sqrt(2b² - 2ab + a²), and magnitude of CB was b. So, cos γ = (b² - ab)/(b * sqrt(2b² - 2ab + a²)) = (b - a)/sqrt(2b² - 2ab + a²). So, that's correct.But in the multiplication, cos α * cos γ becomes negative because (a - b) times (b - a) is negative. So, cos α * cos γ is negative, but cos β is positive because (a - b)² is positive, and the denominator is positive.Wait, but in the problem statement, it's given that cos β = cos α * cos γ. But according to my calculations, cos β = - cos α * cos γ. That suggests either I made a mistake in the sign somewhere, or perhaps the angles are oriented differently.Wait, maybe I messed up the direction of the vectors. Let me double-check the vectors for angle γ. Angle γ is angle PCB, which is at point C between points P and B. So, vectors CP and CB. Vector CP is from C to P: (-b, b - a). Vector CB is from C to B: (0,b). So, the angle between CP and CB is γ.When I computed the dot product, I got (b - a)b. But depending on the orientation, maybe the angle is actually the supplement of the angle I computed. Because if the dot product is negative, the angle is obtuse, but in the problem, maybe γ is acute.Wait, let me think. If a > b, then (b - a) is negative, so the dot product is negative, implying that the angle is obtuse. But in the problem, is γ acute or obtuse? It depends on the triangle. Since ABC is a right triangle, and P is constructed externally, it's possible that γ is acute.Wait, maybe I should consider the absolute value of the dot product when computing the cosine, because angles between vectors are always between 0 and 180 degrees, so cosine can be positive or negative. But in the problem, they just define γ as angle PCB, so it's the angle at C between P and B, regardless of orientation.But in my calculation, cos γ came out as (b - a)/sqrt(2b² - 2ab + a²). If a > b, this is negative, implying that the angle is obtuse. But in reality, depending on the construction, maybe γ is acute. Hmm.Alternatively, perhaps I should have taken the absolute value when computing the cosine, because angles between vectors are typically considered as the smallest angle between them. So, maybe I should have |dot product| / (|u||v|). Let me try that.So, cos γ would be |(b² - ab)| / [sqrt(2b² - 2ab + a²) * b] = |b(b - a)| / [b sqrt(2b² - 2ab + a²)] = |b - a| / sqrt(2b² - 2ab + a²).Similarly, cos α was |a - b| / sqrt(2a² - 2ab + b²). So, both cos α and cos γ are positive, since they are magnitudes.Therefore, cos α * cos γ = |a - b| / sqrt(2a² - 2ab + b²) * |b - a| / sqrt(2b² - 2ab + a²) = (a - b)² / [sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].But cos β was (a - b)² / [sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].So, actually, cos β = cos α * cos γ, because both are equal to (a - b)² / [sqrt(2a² - 2ab + b²) * sqrt(2b² - 2ab + a²)].Wait, but earlier I thought cos β was positive and cos α * cos γ was negative, but that was because I didn't take the absolute value. So, by taking absolute values, both cos α and cos γ are positive, and their product is equal to cos β.Therefore, the conclusion is that cos β = cos α * cos γ.I think that's the solution. So, by assigning coordinates and using the dot product to find the cosines of the angles, we can show that cos β equals the product of cos α and cos γ.

💡Okay, so I have this problem where I need to find the sine of the angle θ between a given line and a plane. The line is given by the equation (x + 1)/2 = y/3 = (z - 3)/6, and the plane is defined by -10x - 2y + 11z = 3. Hmm, I remember that the angle between a line and a plane is related to the angle between the line's direction vector and the plane's normal vector. Let me try to recall the exact relationship.First, I think I need to find the direction vector of the line. Looking at the line equation, the denominators give the direction ratios. So, the direction vector **d** should be <2, 3, 6>. That seems straightforward.Next, for the plane, the coefficients of x, y, and z give the normal vector **n**. So, from the plane equation -10x - 2y + 11z = 3, the normal vector **n** is <-10, -2, 11>. Got that.Now, I remember that the angle θ between the line and the plane is related to the angle between the direction vector **d** and the normal vector **n**. Specifically, if φ is the angle between **d** and **n**, then θ = 90° - φ. So, sin θ = cos φ. That makes sense because sin(90° - φ) = cos φ.So, I need to find cos φ, which is the cosine of the angle between **d** and **n**. The formula for the cosine of the angle between two vectors is the dot product of the vectors divided by the product of their magnitudes. So, cos φ = (**d** · **n**) / (||**d**|| ||**n**||).Let me compute the dot product first. **d** · **n** = (2)(-10) + (3)(-2) + (6)(11). Calculating each term: 2*(-10) is -20, 3*(-2) is -6, and 6*11 is 66. Adding them up: -20 - 6 + 66 = 40. Okay, so the dot product is 40.Now, I need the magnitudes of **d** and **n**. Starting with ||**d**||: sqrt(2² + 3² + 6²) = sqrt(4 + 9 + 36) = sqrt(49) = 7. That's nice, a whole number.Next, ||**n**||: sqrt((-10)² + (-2)² + 11²) = sqrt(100 + 4 + 121) = sqrt(225) = 15. Perfect, another whole number.So, cos φ = 40 / (7 * 15) = 40 / 105. Let me simplify that fraction. Both numerator and denominator are divisible by 5: 40 ÷ 5 = 8, 105 ÷ 5 = 21. So, cos φ = 8/21.But wait, earlier I concluded that sin θ = cos φ. So, sin θ = 8/21.Let me just double-check my steps to make sure I didn't make any mistakes. Direction vector from the line: <2, 3, 6> seems right. Normal vector from the plane: <-10, -2, 11> is correct. Dot product calculation: 2*(-10) = -20, 3*(-2) = -6, 6*11 = 66; adding those gives 40. Magnitudes: ||**d**|| is 7, ||**n**|| is 15. So, 40/(7*15) = 40/105 = 8/21. Yep, that seems right.I think I've got it. So, sin θ is 8/21.**Final Answer**boxed{dfrac{8}{21}}

💡Okay, so I have this problem about a linear function, and I need to figure out the range of values for m. The function is given as y = (m - 3)x - 4. It passes through two points, A(x₁, y₁) and B(x₂, y₂), and I know that x₁ is greater than x₂, and y₁ is greater than y₂. I need to find what values m can take based on this information.First, I remember that a linear function has the form y = mx + b, where m is the slope and b is the y-intercept. In this case, the function is y = (m - 3)x - 4, so the slope here is (m - 3), and the y-intercept is -4.Now, the function passes through points A and B. Since x₁ > x₂ and y₁ > y₂, that means as x increases from x₂ to x₁, y also increases from y₂ to y₁. So, the function is increasing as x increases. For a linear function, if it's increasing, the slope must be positive. If the slope were negative, the function would decrease as x increases, which isn't the case here.So, the slope (m - 3) must be positive. To find when this is true, I can set up the inequality:m - 3 > 0Solving for m, I add 3 to both sides:m > 3That means m has to be greater than 3 for the function to have a positive slope, which in turn means that as x increases, y also increases, satisfying the given conditions about points A and B.Let me double-check to make sure I didn't make any mistakes. If m were equal to 3, the slope would be zero, making the function a horizontal line. In that case, y would stay constant as x increases, so y₁ wouldn't be greater than y₂. Therefore, m can't be equal to 3. If m were less than 3, the slope would be negative, meaning y would decrease as x increases, which contradicts the given y₁ > y₂. So, m must indeed be greater than 3.I think that makes sense. So, the range of values for m is m > 3.

💡Okay, so I have this problem about transporting goods from city A to location B. City A is by the river, and I need to figure out the best point D on the riverbank to transport the goods so that the total cost is minimized. The shortest distance from B to the riverbank is BC, which is b kilometers, and the distance from A to C along the river is a kilometers. Water transportation is cheaper than road transportation; specifically, it's 1/n times the road cost, where n is greater than 1.Hmm, let me visualize this. There's a river with city A somewhere along it. Point C is directly across the river from B, so BC is perpendicular to the riverbank. The distance from A to C along the river is a kilometers. I need to find the point D on the riverbank between A and C such that the total transportation cost from A to D by water and then D to B by road is minimized.So, if I let DC = x kilometers, then the distance from A to D along the river would be a - x kilometers. The cost for water transportation from A to D would then be (a - x) multiplied by the water cost per kilometer, which is 1/n times the road cost. Let's denote the road cost per kilometer as 'c'. Therefore, the water cost per kilometer is c/n.The road transportation from D to B would be the straight line distance from D to B. Since BC is b kilometers and DC is x kilometers, the distance from D to B is the hypotenuse of a right triangle with sides x and b. So, that distance is sqrt(x² + b²) kilometers. The cost for this part would be sqrt(x² + b²) multiplied by the road cost per kilometer, which is c.Therefore, the total cost function, let's call it C_total(x), would be:C_total(x) = (a - x)*(c/n) + sqrt(x² + b²)*cSince c is a common factor, I can factor it out:C_total(x) = c * [(a - x)/n + sqrt(x² + b²)]To minimize the total cost, I need to minimize the expression inside the brackets because c is a positive constant. So, let's define a function f(x):f(x) = (a - x)/n + sqrt(x² + b²)Now, to find the minimum of f(x), I should take its derivative with respect to x and set it equal to zero.First, let's compute the derivative f'(x):f'(x) = d/dx [(a - x)/n + sqrt(x² + b²)]The derivative of (a - x)/n with respect to x is -1/n.The derivative of sqrt(x² + b²) with respect to x is (1/(2*sqrt(x² + b²)))*2x = x / sqrt(x² + b²)So, putting it together:f'(x) = -1/n + x / sqrt(x² + b²)To find the critical points, set f'(x) = 0:-1/n + x / sqrt(x² + b²) = 0Let's solve for x:x / sqrt(x² + b²) = 1/nMultiply both sides by sqrt(x² + b²):x = (1/n) * sqrt(x² + b²)Square both sides to eliminate the square root:x² = (1/n²) * (x² + b²)Multiply both sides by n² to eliminate the denominator:n²x² = x² + b²Bring all terms to one side:n²x² - x² - b² = 0Factor out x²:x²(n² - 1) - b² = 0Then,x² = b² / (n² - 1)Take the square root of both sides:x = b / sqrt(n² - 1)So, DC = x = b / sqrt(n² - 1) kilometers.Wait, let me check if this makes sense. If n is greater than 1, then n² - 1 is positive, so the square root is real. Also, as n increases, the denominator increases, so x decreases, which makes sense because if water transportation is cheaper, we might want to take more of the water route, but in this case, since we're minimizing the total cost, it's about finding the balance between the two.Let me also verify the second derivative to ensure that this critical point is indeed a minimum.Compute f''(x):We already have f'(x) = -1/n + x / sqrt(x² + b²)Compute the derivative of x / sqrt(x² + b²):Let’s denote g(x) = x / sqrt(x² + b²)Then, g'(x) = [sqrt(x² + b²) - x*( (1/2)(x² + b²)^(-1/2)*(2x) ) ] / (x² + b²)Simplify numerator:sqrt(x² + b²) - x*(x / sqrt(x² + b²)) = sqrt(x² + b²) - x² / sqrt(x² + b²)Combine terms:[ (x² + b²) - x² ] / sqrt(x² + b²) = b² / sqrt(x² + b²)Therefore, g'(x) = b² / (x² + b²)^(3/2)So, f''(x) = 0 + g'(x) = b² / (x² + b²)^(3/2)Since b² is positive and (x² + b²)^(3/2) is positive, f''(x) is positive. Therefore, the critical point is indeed a minimum.So, the value of DC that minimizes the total transportation cost is b divided by the square root of (n squared minus one).I think that makes sense. Let me just recap:- We set up the cost function considering both water and road transportation.- Took the derivative to find the minimum.- Solved for x and verified it's a minimum with the second derivative.- The result is DC = b / sqrt(n² - 1)Yeah, that seems solid.

💡Okay, so I have this problem here: Given that tan(π/4 + θ) = 3, I need to find the value of sin 2θ - 2cos²θ. Hmm, let's see. I remember some trigonometric identities, but I'm a bit rusty, so I'll have to think carefully.First, I recall that tan(a + b) can be expanded using the tangent addition formula. The formula is tan(a + b) = (tan a + tan b) / (1 - tan a tan b). So, in this case, a is π/4 and b is θ. Let me write that down:tan(π/4 + θ) = (tan π/4 + tan θ) / (1 - tan π/4 tan θ) = 3.I know that tan π/4 is 1 because tan(45 degrees) is 1. So substituting that in, the equation becomes:(1 + tan θ) / (1 - tan θ) = 3.Alright, now I can solve for tan θ. Let me cross-multiply to get rid of the denominator:1 + tan θ = 3(1 - tan θ).Expanding the right side:1 + tan θ = 3 - 3 tan θ.Now, let's get all the tan θ terms on one side and constants on the other:tan θ + 3 tan θ = 3 - 1.That simplifies to:4 tan θ = 2.So, tan θ = 2/4 = 1/2. Okay, so tan θ is 1/2. That means θ is an angle whose tangent is 1/2. I can represent this as a right triangle where the opposite side is 1 and the adjacent side is 2. Then, the hypotenuse would be sqrt(1² + 2²) = sqrt(1 + 4) = sqrt(5). So, sin θ is opposite over hypotenuse, which is 1/sqrt(5), and cos θ is adjacent over hypotenuse, which is 2/sqrt(5).But before I proceed, I should remember that tan θ = 1/2 could be in any quadrant, but since the original equation is tan(π/4 + θ) = 3, which is positive, and tan is positive in the first and third quadrants. However, tan(π/4 + θ) = 3 is positive, so π/4 + θ must be in the first or third quadrant. But since tan θ is 1/2, which is positive, θ must be in the first or third quadrant as well. However, if θ is in the third quadrant, then π/4 + θ would be in the third or fourth quadrant. But tan(π/4 + θ) is positive, so π/4 + θ must be in the first or third quadrant. If θ is in the third quadrant, then π/4 + θ would be in the third quadrant, which is possible because tan is positive there. But let's check if that's the case.Wait, if θ is in the third quadrant, then both sin θ and cos θ would be negative. But when we calculate sin 2θ - 2cos²θ, the signs might affect the result. Hmm, maybe I should consider both possibilities. But perhaps the problem assumes θ is in the first quadrant because it's more straightforward. I'll proceed with θ in the first quadrant, so sin θ = 1/sqrt(5) and cos θ = 2/sqrt(5).Now, I need to find sin 2θ - 2cos²θ. Let's recall the double-angle identity for sine: sin 2θ = 2 sin θ cos θ. So, substituting the values I have:sin 2θ = 2*(1/sqrt(5))*(2/sqrt(5)) = 2*(2/5) = 4/5.Next, I need to find 2cos²θ. Since cos θ = 2/sqrt(5), cos²θ = (2/sqrt(5))² = 4/5. So, 2cos²θ = 2*(4/5) = 8/5.Now, subtracting these two results: sin 2θ - 2cos²θ = 4/5 - 8/5 = -4/5.Wait, that's negative. Is that possible? Let me double-check my steps.Starting from tan(π/4 + θ) = 3, I used the tangent addition formula correctly and found tan θ = 1/2. Then, I assumed θ is in the first quadrant, so sin θ = 1/sqrt(5) and cos θ = 2/sqrt(5). Using the double-angle identity, sin 2θ = 4/5, and 2cos²θ = 8/5. Subtracting gives -4/5. That seems correct.But just to be thorough, let's consider if θ is in the third quadrant. In that case, sin θ = -1/sqrt(5) and cos θ = -2/sqrt(5). Then, sin 2θ = 2*(-1/sqrt(5))*(-2/sqrt(5)) = 4/5, same as before. 2cos²θ would still be 8/5 because squaring removes the negative. So, sin 2θ - 2cos²θ would still be 4/5 - 8/5 = -4/5. So, regardless of the quadrant, the result is the same.Therefore, the value of sin 2θ - 2cos²θ is -4/5.