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If cos A=(28)/(53) and tan B=(5)/(12) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If cos⁡A=2853 \cos A=\frac{28}{53} and tan⁡B=512 \tan B=\frac{5}{12} and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer:

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Q. If cos⁡A=2853 \cos A=\frac{28}{53} and tan⁡B=512 \tan B=\frac{5}{12} and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer:
  1. Find tan⁡A\tan A: Use the identity for the tangent of a sum of two angles: tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}. First, we need to find tan⁡A\tan A using the given cos⁡A=2853\cos A = \frac{28}{53}. Since cos⁡A=adjacenthypotenuse\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}, we can find the opposite side using the Pythagorean theorem: opposite2=hypotenuse2−adjacent2\text{opposite}^2 = \text{hypotenuse}^2 - \text{adjacent}^2.
  2. Calculate opposite side: Calculate the opposite side for angle AA: opposite2=532−282.\text{opposite}^2 = 53^2 - 28^2.opposite2=2809−784.\text{opposite}^2 = 2809 - 784.opposite2=2025.\text{opposite}^2 = 2025.opposite=2025.\text{opposite} = \sqrt{2025}.opposite=45.\text{opposite} = 45.
  3. Find tan⁡B\tan B: Now we can find tan⁡A\tan A: tan⁡A=oppositeadjacent.\tan A = \frac{\text{opposite}}{\text{adjacent}}.\newlinetan⁡A=4528.\tan A = \frac{45}{28}.
  4. Apply tan⁡(A+B)\tan(A+B) identity: We already have tan⁡B\tan B given as 512\frac{5}{12}.\newlineNow we can use the identity for tan⁡(A+B)\tan(A+B): tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}.\newlinetan⁡(A+B)=4528+5121−(4528⋅512)\tan(A+B) = \frac{\frac{45}{28} + \frac{5}{12}}{1 - \left(\frac{45}{28} \cdot \frac{5}{12}\right)}.
  5. Add fractions: Find a common denominator for adding the fractions 4528\frac{45}{28} and 512\frac{5}{12}. The common denominator is 8484. Convert the fractions: (4528)×(33)=13584\left(\frac{45}{28}\right) \times \left(\frac{3}{3}\right) = \frac{135}{84} and (512)×(77)=3584\left(\frac{5}{12}\right) \times \left(\frac{7}{7}\right) = \frac{35}{84}. Now add the fractions: 13584+3584=17084\frac{135}{84} + \frac{35}{84} = \frac{170}{84}.
  6. Simplify fraction: Simplify the fraction 17084\frac{170}{84} by dividing both numerator and denominator by 22.\newline17084=8542\frac{170}{84} = \frac{85}{42}.
  7. Calculate denominator: Now calculate the denominator of the tan⁡(A+B)\tan(A+B) formula: 1−(4528×512)1 - \left(\frac{45}{28} \times \frac{5}{12}\right). First, multiply the fractions: (4528)×(512)=225336\left(\frac{45}{28}\right) \times \left(\frac{5}{12}\right) = \frac{225}{336}. Simplify the fraction 225336\frac{225}{336} by dividing both numerator and denominator by 33. 225336=75112.\frac{225}{336} = \frac{75}{112}.
  8. Subtract fraction from 11: Now subtract the fraction from 11: 1−751121 - \frac{75}{112}. Convert 11 to a fraction with the same denominator: 112112−75112=37112\frac{112}{112} - \frac{75}{112} = \frac{37}{112}.
  9. Calculate tan⁡(A+B)\tan(A+B) formula: Now we have both the numerator and the denominator for the tan⁡(A+B)\tan(A+B) formula.\newlinetan⁡(A+B)=8542/37112\tan(A+B) = \frac{85}{42} / \frac{37}{112}.\newlineTo divide by a fraction, multiply by its reciprocal: 8542×11237\frac{85}{42} \times \frac{112}{37}.
  10. Multiply fractions: Multiply the fractions: (85×112)/(42×37)(85 \times 112) / (42 \times 37).85×112=952085 \times 112 = 9520 and 42×37=155442 \times 37 = 1554.tan⁡(A+B)=95201554\tan(A+B) = \frac{9520}{1554}.
  11. Simplify fraction: Simplify the fraction 95201554\frac{9520}{1554} by finding the greatest common divisor (GCD) and dividing both numerator and denominator by it.\newlineThe GCD of 95209520 and 15541554 is 22.\newline95202=4760\frac{9520}{2} = 4760 and 15542=777\frac{1554}{2} = 777.\newlinetan⁡(A+B)=4760777\tan(A+B) = \frac{4760}{777}.
  12. Simplify further: Simplify the fraction 4760777\frac{4760}{777} further if possible.\newlineThe GCD of 47604760 and 777777 is 11, so the fraction is already in its simplest form.\newlinetan⁡(A+B)=4760777\tan(A+B) = \frac{4760}{777}.

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