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Advanced Trig Identities - IB Mathematics Analysis and Approaches

  1. Given cos⁡(x+y)=14\cos(x+y) = \frac{1}{4} and cos⁡(x−y)=34\cos(x-y) = \frac{3}{4}, what is the value of cos⁡(x)cos⁡(y)\cos(x)\cos(y)?
    Answer: 12\frac{1}{2} We use the product-to-sum identity cos⁡(x)cos⁡(y)=12[cos⁡(x+y)+cos⁡(x−y)]\cos(x)\cos(y) = \frac{1}{2}[\cos(x+y) + \cos(x-y)]. Substituting the given values directly into this formula: cos⁡(x)cos⁡(y)=12[14+34]=12[44]=12(1)=12\cos(x)\cos(y) = \frac{1}{2}\left[\frac{1}{4} + \frac{3}{4}\right] = \frac{1}{2}\left[\frac{4}{4}\right] = \frac{1}{2}(1) = \frac{1}{2}.
  2. Which of the following is an indefinite integral of f(x)=4sin⁡(x)cos⁡(3x)f(x) = 4\sin(x)\cos(3x)?
    Answer: cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C First, use the product-to-sum identity sin⁡(A)cos⁡(B)=12[sin⁡(A+B)+sin⁡(A−B)]\sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)]. Let A=xA=x and B=3xB=3x. Then 4sin⁡(x)cos⁡(3x)=4⋅12[sin⁡(x+3x)+sin⁡(x−3x)]=2[sin⁡(4x)+sin⁡(−2x)]=2sin⁡(4x)−2sin⁡(2x)4\sin(x)\cos(3x) = 4 \cdot \frac{1}{2}[\sin(x+3x) + \sin(x-3x)] = 2[\sin(4x) + \sin(-2x)] = 2\sin(4x) - 2\sin(2x). Now, integrate this expression: ∫(2sin⁡(4x)−2sin⁡(2x))dx=2(−cos⁡(4x)4)−2(−cos⁡(2x)2)+C=−12cos⁡(4x)+cos⁡(2x)+C\int (2\sin(4x) - 2\sin(2x)) dx = 2\left(-\frac{\cos(4x)}{4}\right) - 2\left(-\frac{\cos(2x)}{2}\right) + C = -\frac{1}{2}\cos(4x) + \cos(2x) + C, which can be written as cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C.
  3. If tan⁡(A)=12\tan(A) = \frac{1}{2} and tan⁡(A+B)=3\tan(A+B) = 3, what is the value of tan⁡(B)\tan(B)?
    Answer: 1 Start with the compound angle identity for tangent: tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}. Substitute the given values: 3=12+tan⁡(B)1−12tan⁡(B)3 = \frac{\frac{1}{2} + \tan(B)}{1 - \frac{1}{2}\tan(B)}. Let t=tan⁡(B)t = \tan(B). Then 3(1−12t)=12+t3(1 - \frac{1}{2}t) = \frac{1}{2} + t. This simplifies to 3−32t=12+t3 - \frac{3}{2}t = \frac{1}{2} + t. Rearranging to solve for tt: 3−12=t+32t3 - \frac{1}{2} = t + \frac{3}{2}t, which gives 52=52t\frac{5}{2} = \frac{5}{2}t. Therefore, t=tan⁡(B)=1t = \tan(B) = 1.
  4. Which of the following expressions is identically equal to cos⁡(x)−cos⁡(3x)sin⁡(3x)−sin⁡(x)\frac{\cos(x) - \cos(3x)}{\sin(3x) - \sin(x)}?
    Answer: tan⁡(2x)\tan(2x) We use the sum-to-product identities. For the numerator: cos⁡(P)−cos⁡(Q)=−2sin⁡(P+Q2)sin⁡(P−Q2)\cos(P) - \cos(Q) = -2\sin\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right). So, cos⁡(x)−cos⁡(3x)=−2sin⁡(x+3x2)sin⁡(x−3x2)=−2sin⁡(2x)sin⁡(−x)=2sin⁡(2x)sin⁡(x)\cos(x) - \cos(3x) = -2\sin\left(\frac{x+3x}{2}\right)\sin\left(\frac{x-3x}{2}\right) = -2\sin(2x)\sin(-x) = 2\sin(2x)\sin(x). For the denominator: sin⁡(P)−sin⁡(Q)=2cos⁡(P+Q2)sin⁡(P−Q2)\sin(P) - \sin(Q) = 2\cos\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right). So, sin⁡(3x)−sin⁡(x)=2cos⁡(3x+x2)sin⁡(3x−x2)=2cos⁡(2x)sin⁡(x)\sin(3x) - \sin(x) = 2\cos\left(\frac{3x+x}{2}\right)\sin\left(\frac{3x-x}{2}\right) = 2\cos(2x)\sin(x). The fraction becomes 2sin⁡(2x)sin⁡(x)2cos⁡(2x)sin⁡(x)=sin⁡(2x)cos⁡(2x)=tan⁡(2x)\frac{2\sin(2x)\sin(x)}{2\cos(2x)\sin(x)} = \frac{\sin(2x)}{\cos(2x)} = \tan(2x).
  5. Which expression is equivalent to sin⁡(x)+sin⁡(5x)cos⁡(x)+cos⁡(5x)\frac{\sin(x) + \sin(5x)}{\cos(x) + \cos(5x)}?
    Answer: tan⁡(3x)\tan(3x) Apply the sum-to-product identities. Numerator: sin⁡(x)+sin⁡(5x)=2sin⁡(x+5x2)cos⁡(x−5x2)=2sin⁡(3x)cos⁡(−2x)=2sin⁡(3x)cos⁡(2x)\sin(x) + \sin(5x) = 2\sin\left(\frac{x+5x}{2}\right)\cos\left(\frac{x-5x}{2}\right) = 2\sin(3x)\cos(-2x) = 2\sin(3x)\cos(2x). Denominator: cos⁡(x)+cos⁡(5x)=2cos⁡(x+5x2)cos⁡(x−5x2)=2cos⁡(3x)cos⁡(−2x)=2cos⁡(3x)cos⁡(2x)\cos(x) + \cos(5x) = 2\cos\left(\frac{x+5x}{2}\right)\cos\left(\frac{x-5x}{2}\right) = 2\cos(3x)\cos(-2x) = 2\cos(3x)\cos(2x). The fraction becomes 2sin⁡(3x)cos⁡(2x)2cos⁡(3x)cos⁡(2x)=sin⁡(3x)cos⁡(3x)=tan⁡(3x)\frac{2\sin(3x)\cos(2x)}{2\cos(3x)\cos(2x)} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x).
  6. In triangle ABC, sin⁡(A)=35\sin(A) = \frac{3}{5} and sin⁡(B)=513\sin(B) = \frac{5}{13}. Given that angle A is acute and angle B is obtuse, find cos⁡(C)\cos(C).
    Answer: 6365\frac{63}{65} Since angle A is acute, cos⁡(A)>0\cos(A) > 0. cos⁡(A)=1−(3/5)2=45\cos(A) = \sqrt{1 - (3/5)^2} = \frac{4}{5}. Since angle B is obtuse, cos⁡(B)<0\cos(B) < 0. cos⁡(B)=−1−(5/13)2=−1213\cos(B) = -\sqrt{1 - (5/13)^2} = -\frac{12}{13}. In a triangle, A+B+C=πA+B+C = \pi, so C=π−(A+B)C = \pi - (A+B). Therefore, cos⁡(C)=cos⁡(π−(A+B))=−cos⁡(A+B)\cos(C) = \cos(\pi - (A+B)) = -\cos(A+B). Using the compound angle identity, cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)=(45)(−1213)−(35)(513)=−4865−1565=−6365\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) = \left(\frac{4}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) = -\frac{48}{65} - \frac{15}{65} = -\frac{63}{65}. Finally, cos⁡(C)=−(−6365)=6365\cos(C) = -(-\frac{63}{65}) = \frac{63}{65}.
  7. Find the value of cos⁡(4θ)\cos(4\theta) if cos⁡(θ)=13\cos(\theta) = \frac{1}{3}.
    Answer: 1781\frac{17}{81} We can find cos⁡(4θ)\cos(4\theta) by applying the double angle identity for cosine twice. First, find cos⁡(2θ)\cos(2\theta): cos⁡(2θ)=2cos⁡2(θ)−1=2(13)2−1=2(19)−1=29−1=−79\cos(2\theta) = 2\cos^2(\theta) - 1 = 2\left(\frac{1}{3}\right)^2 - 1 = 2\left(\frac{1}{9}\right) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}. Now, use this result to find cos⁡(4θ)=cos⁡(2(2θ))\cos(4\theta) = \cos(2(2\theta)): cos⁡(4θ)=2cos⁡2(2θ)−1=2(−79)2−1=2(4981)−1=9881−8181=1781\cos(4\theta) = 2\cos^2(2\theta) - 1 = 2\left(-\frac{7}{9}\right)^2 - 1 = 2\left(\frac{49}{81}\right) - 1 = \frac{98}{81} - \frac{81}{81} = \frac{17}{81}.
  8. Find the smallest positive solution to the equation sin⁡(4x)cos⁡(2x)−cos⁡(4x)sin⁡(2x)=12\sin(4x)\cos(2x) - \cos(4x)\sin(2x) = \frac{1}{2}.
    Answer: π12\frac{\pi}{12} The left side of the equation matches the compound angle identity sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B), with A=4xA=4x and B=2xB=2x. So, the equation simplifies to sin⁡(4x−2x)=sin⁡(2x)=12\sin(4x - 2x) = \sin(2x) = \frac{1}{2}. The principal value for 2x2x is arcsin⁡(12)=π6\arcsin(\frac{1}{2}) = \frac{\pi}{6}. The general solutions for 2x2x are 2x=π6+2nπ2x = \frac{\pi}{6} + 2n\pi and 2x=π−π6+2nπ=5π6+2nπ2x = \pi - \frac{\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi. This gives x=π12+nπx = \frac{\pi}{12} + n\pi and x=5π12+nπx = \frac{5\pi}{12} + n\pi. The smallest positive solution occurs when n=0n=0, which is x=π12x = \frac{\pi}{12}.
  9. The expression 1−cos⁡(2x)+sin⁡(2x)1+cos⁡(2x)+sin⁡(2x)\frac{1 - \cos(2x) + \sin(2x)}{1 + \cos(2x) + \sin(2x)} simplifies to which of the following?
    Answer: tan⁡(x)\tan(x) Use the double angle identities to express everything in terms of xx. Use cos⁡(2x)=1−2sin⁡2(x)\cos(2x) = 1 - 2\sin^2(x) for the numerator and cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1 for the denominator, along with sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x). Numerator: 1−(1−2sin⁡2(x))+2sin⁡(x)cos⁡(x)=2sin⁡2(x)+2sin⁡(x)cos⁡(x)=2sin⁡(x)(sin⁡(x)+cos⁡(x))1 - (1 - 2\sin^2(x)) + 2\sin(x)\cos(x) = 2\sin^2(x) + 2\sin(x)\cos(x) = 2\sin(x)(\sin(x) + \cos(x)). Denominator: 1+(2cos⁡2(x)−1)+2sin⁡(x)cos⁡(x)=2cos⁡2(x)+2sin⁡(x)cos⁡(x)=2cos⁡(x)(cos⁡(x)+sin⁡(x))1 + (2\cos^2(x) - 1) + 2\sin(x)\cos(x) = 2\cos^2(x) + 2\sin(x)\cos(x) = 2\cos(x)(\cos(x) + \sin(x)). The fraction becomes 2sin⁡(x)(sin⁡(x)+cos⁡(x))2cos⁡(x)(cos⁡(x)+sin⁡(x))=sin⁡(x)cos⁡(x)=tan⁡(x)\frac{2\sin(x)(\sin(x) + \cos(x))}{2\cos(x)(\cos(x) + \sin(x))} = \frac{\sin(x)}{\cos(x)} = \tan(x), assuming sin⁡(x)+cos⁡(x)≠0\sin(x) + \cos(x) \neq 0 and cos⁡(x)≠0\cos(x) \neq 0.
  10. Simplify the expression 2sin⁡(5x)cos⁡(3x)−sin⁡(8x)2\sin(5x)\cos(3x) - \sin(8x).
    Answer: sin⁡(2x)\sin(2x) Use the product-to-sum identity 2sin⁡(A)cos⁡(B)=sin⁡(A+B)+sin⁡(A−B)2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B). Let A=5xA=5x and B=3xB=3x. Then, 2sin⁡(5x)cos⁡(3x)=sin⁡(5x+3x)+sin⁡(5x−3x)=sin⁡(8x)+sin⁡(2x)2\sin(5x)\cos(3x) = \sin(5x+3x) + \sin(5x-3x) = \sin(8x) + \sin(2x). The original expression becomes (sin⁡(8x)+sin⁡(2x))−sin⁡(8x)=sin⁡(2x)(\sin(8x) + \sin(2x)) - \sin(8x) = \sin(2x).
  11. What is the maximum value of the function f(x)=sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x)f(x) = \sin(x)\cos(x) + \cos^2(x) - \sin^2(x)?
    Answer: 52\frac{\sqrt{5}}{2} The function can be rewritten using double angle identities. We know that sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x) and cos⁡2(x)−sin⁡2(x)=cos⁡(2x)\cos^2(x) - \sin^2(x) = \cos(2x). So, f(x)=12sin⁡(2x)+cos⁡(2x)f(x) = \frac{1}{2}\sin(2x) + \cos(2x). This is in the form asin⁡(θ)+bcos⁡(θ)a\sin(\theta) + b\cos(\theta) where θ=2x\theta = 2x, a=1/2a = 1/2, and b=1b = 1. The maximum value of this expression is R=a2+b2R = \sqrt{a^2 + b^2}. In this case, R=(12)2+12=14+1=54=52R = \sqrt{(\frac{1}{2})^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.
  12. Given that sin⁡(A)=35\sin(A) = \frac{3}{5} for π2<A<π\frac{\pi}{2} < A < \pi and cos⁡(B)=−1213\cos(B) = -\frac{12}{13} for π<B<3π2\pi < B < \frac{3\pi}{2}, find the value of cos⁡(A+B)\cos(A+B).
    Answer: 6365\frac{63}{65} First, determine the values of cos⁡(A)\cos(A) and sin⁡(B)\sin(B). Since AA is in Quadrant II, cos⁡(A)\cos(A) is negative. Using sin⁡2(A)+cos⁡2(A)=1\sin^2(A) + \cos^2(A) = 1, we have cos⁡(A)=−1−(3/5)2=−1−9/25=−16/25=−45\cos(A) = -\sqrt{1 - (3/5)^2} = -\sqrt{1 - 9/25} = -\sqrt{16/25} = -\frac{4}{5}. Since BB is in Quadrant III, sin⁡(B)\sin(B) is negative. Using sin⁡2(B)+cos⁡2(B)=1\sin^2(B) + \cos^2(B) = 1, we have sin⁡(B)=−1−(−12/13)2=−1−144/169=−25/169=−513\sin(B) = -\sqrt{1 - (-12/13)^2} = -\sqrt{1 - 144/169} = -\sqrt{25/169} = -\frac{5}{13}. Now, use the compound angle identity for cosine: cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B). Substituting the values: cos⁡(A+B)=(−45)(−1213)−(35)(−513)=4865+1565=6365\cos(A+B) = \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(-\frac{5}{13}\right) = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}.
  13. What is the period of the function f(x)=cos⁡(4x)cos⁡(2x)−sin⁡(4x)sin⁡(2x)f(x) = \cos(4x)\cos(2x) - \sin(4x)\sin(2x)?
    Answer: π3\frac{\pi}{3} The expression cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A)\cos(B) - \sin(A)\sin(B) is the compound angle identity for cos⁡(A+B)\cos(A+B). In this case, A=4xA=4x and B=2xB=2x. Therefore, f(x)=cos⁡(4x+2x)=cos⁡(6x)f(x) = \cos(4x+2x) = \cos(6x). The period of a function g(x)=acos⁡(kx)+dg(x) = a\cos(kx)+d is given by 2π∣k∣\frac{2\pi}{|k|}. For f(x)=cos⁡(6x)f(x) = \cos(6x), we have k=6k=6. The period is 2π6=π3\frac{2\pi}{6} = \frac{\pi}{3}.
  14. Find the exact value of sin⁡(7π12)+sin⁡(π12)\sin(\frac{7\pi}{12}) + \sin(\frac{\pi}{12}).
    Answer: 62\frac{\sqrt{6}}{2} Use the sum-to-product identity sin⁡(P)+sin⁡(Q)=2sin⁡(P+Q2)cos⁡(P−Q2)\sin(P) + \sin(Q) = 2\sin\left(\frac{P+Q}{2}\right)\cos\left(\frac{P-Q}{2}\right). Here, P=7π12P = \frac{7\pi}{12} and Q=π12Q = \frac{\pi}{12}. So P+Q2=8π/122=2π/32=π3\frac{P+Q}{2} = \frac{8\pi/12}{2} = \frac{2\pi/3}{2} = \frac{\pi}{3}, and P−Q2=6π/122=π/22=π4\frac{P-Q}{2} = \frac{6\pi/12}{2} = \frac{\pi/2}{2} = \frac{\pi}{4}. The expression becomes 2sin⁡(π3)cos⁡(π4)=2⋅32⋅22=264=622\sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}.
  15. The expression sin⁡(3x)\sin(3x) is equivalent to which of the following?
    Answer: 3sin⁡(x)−4sin⁡3(x)3\sin(x) - 4\sin^3(x) To derive the identity for sin⁡(3x)\sin(3x), we use the compound angle and double angle formulas: sin⁡(3x)=sin⁡(2x+x)=sin⁡(2x)cos⁡(x)+cos⁡(2x)sin⁡(x)=(2sin⁡(x)cos⁡(x))cos⁡(x)+(1−2sin⁡2(x))sin⁡(x)=2sin⁡(x)cos⁡2(x)+sin⁡(x)−2sin⁡3(x)=2sin⁡(x)(1−sin⁡2(x))+sin⁡(x)−2sin⁡3(x)=2sin⁡(x)−2sin⁡3(x)+sin⁡(x)−2sin⁡3(x)=3sin⁡(x)−4sin⁡3(x)\sin(3x) = \sin(2x+x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) = (2\sin(x)\cos(x))\cos(x) + (1-2\sin^2(x))\sin(x) = 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^3(x) = 2\sin(x)(1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 3\sin(x) - 4\sin^3(x).
  16. Given cos⁡(x)=725\cos(x) = \frac{7}{25} and 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, find the value of cos⁡(x2)\cos(\frac{x}{2}).
    Answer: −45-\frac{4}{5} Use the half-angle identity derived from cos⁡(2A)=2cos⁡2(A)−1\cos(2A) = 2\cos^2(A) - 1. Let A=x/2A = x/2, so cos⁡(x)=2cos⁡2(x2)−1\cos(x) = 2\cos^2(\frac{x}{2}) - 1. Rearranging gives cos⁡2(x2)=1+cos⁡(x)2\cos^2(\frac{x}{2}) = \frac{1+\cos(x)}{2}. Substituting the value of cos⁡(x)\cos(x): cos⁡2(x2)=1+7/252=32/252=1625\cos^2(\frac{x}{2}) = \frac{1 + 7/25}{2} = \frac{32/25}{2} = \frac{16}{25}. Thus, cos⁡(x2)=±1625=±45\cos(\frac{x}{2}) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}. To determine the sign, consider the quadrant of x2\frac{x}{2}. Since 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, dividing by 2 gives 3π4<x2<π\frac{3\pi}{4} < \frac{x}{2} < \pi. This means x2\frac{x}{2} is in Quadrant II, where cosine is negative. Therefore, cos⁡(x2)=−45\cos(\frac{x}{2}) = -\frac{4}{5}.
  17. Given that sin⁡(x)=−25\sin(x) = -\frac{2}{\sqrt{5}} and π<x<3π2\pi < x < \frac{3\pi}{2}, find tan⁡(2x)\tan(2x).
    Answer: −43-\frac{4}{3} First, find cos⁡(x)\cos(x). Since xx is in Quadrant III, cos⁡(x)\cos(x) is negative. cos⁡(x)=−1−sin⁡2(x)=−1−(−25)2=−1−45=−15=−15\cos(x) = -\sqrt{1 - \sin^2(x)} = -\sqrt{1 - \left(-\frac{2}{\sqrt{5}}\right)^2} = -\sqrt{1 - \frac{4}{5}} = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}}. Next, find tan⁡(x)=sin⁡(x)cos⁡(x)=−2/5−1/5=2\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{-2/\sqrt{5}}{-1/\sqrt{5}} = 2. Now, use the double angle identity for tangent: tan⁡(2x)=2tan⁡(x)1−tan⁡2(x)\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}. Substitute tan⁡(x)=2\tan(x) = 2: tan⁡(2x)=2(2)1−(2)2=41−4=−43\tan(2x) = \frac{2(2)}{1 - (2)^2} = \frac{4}{1 - 4} = -\frac{4}{3}.
  18. If A+B=π4A+B = \frac{\pi}{4}, what is the value of (1+tan⁡A)(1+tan⁡B)(1+\tan A)(1+\tan B)?
    Answer: 2 Since A+B=π4A+B = \frac{\pi}{4}, we have tan⁡(A+B)=tan⁡(π4)=1\tan(A+B) = \tan(\frac{\pi}{4}) = 1. Using the compound angle formula for tangent, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}. So, tan⁡A+tan⁡B1−tan⁡Atan⁡B=1\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1. This implies tan⁡A+tan⁡B=1−tan⁡Atan⁡B\tan A + \tan B = 1 - \tan A \tan B, which can be rearranged to tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A \tan B = 1. Now consider the expression we want to evaluate: (1+tan⁡A)(1+tan⁡B)=1+tan⁡A+tan⁡B+tan⁡Atan⁡B(1+\tan A)(1+\tan B) = 1 + \tan A + \tan B + \tan A \tan B. From our rearrangement, we know that tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A \tan B = 1. Therefore, (1+tan⁡A)(1+tan⁡B)=1+1=2(1+\tan A)(1+\tan B) = 1 + 1 = 2.

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