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If tan A=(12)/(35) and cos B=(15)/(17) and angles A and B are in Quadrant I, find the value of tan(A-B). Answer:

If tan⁡A=1235 \tan A=\frac{12}{35} and cos⁡B=1517 \cos B=\frac{15}{17} 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 tan⁡A=1235 \tan A=\frac{12}{35} and cos⁡B=1517 \cos B=\frac{15}{17} and angles A and B are in Quadrant I, find the value of tan⁡(A−B) \tan (A-B) .\newlineAnswer:
  1. Use tan⁡(A−B)\tan(A-B) Formula: Use the formula for tan⁡(A−B)\tan(A-B), which is 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}. We are given tan⁡A=1235\tan A = \frac{12}{35}. We need to find tan⁡B\tan B using the given cos⁡B=1517\cos B = \frac{15}{17}.
  2. Find tan⁡B\tan B: Since cos⁡B=1517\cos B = \frac{15}{17}, we can find sin⁡B\sin B using the Pythagorean identity sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1.\newlinesin⁡2B=1−cos⁡2B\sin^2 B = 1 - \cos^2 B\newlinesin⁡2B=1−(1517)2\sin^2 B = 1 - \left(\frac{15}{17}\right)^2\newlinesin⁡2B=1−225289\sin^2 B = 1 - \frac{225}{289}\newlinesin⁡2B=289−225289\sin^2 B = \frac{289 - 225}{289}\newlinesin⁡2B=64289\sin^2 B = \frac{64}{289}\newlinesin⁡B=64289\sin B = \sqrt{\frac{64}{289}}\newlinecos⁡B=1517\cos B = \frac{15}{17}00, since cos⁡B=1517\cos B = \frac{15}{17}11 is in Quadrant I, sin⁡B\sin B is positive.
  3. Calculate tan⁡B\tan B: Now we can find tan⁡B\tan B using the ratio tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}.tan⁡B=817/1517\tan B = \frac{8}{17} / \frac{15}{17}tan⁡B=817⋅1715\tan B = \frac{8}{17} \cdot \frac{17}{15}tan⁡B=815\tan B = \frac{8}{15}
  4. Substitute tan Values: Substitute the values of tan⁡A\tan A and tan⁡B\tan B into the tan⁡(A−B)\tan(A-B) formula.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}tan⁡(A−B)=(1235)−(815)1+(1235)⋅(815)\tan(A-B) = \frac{(\frac{12}{35}) - (\frac{8}{15})}{1 + (\frac{12}{35}) \cdot (\frac{8}{15})}
  5. Simplify Numerator/Denominator: Simplify the numerator and the denominator separately.\newlineNumerator: (1235)−(815)=(12×3−8×735×3)(\frac{12}{35}) - (\frac{8}{15}) = (\frac{12\times3 - 8\times7}{35\times3})\newlineNumerator: (36−56105)=−20105=−421(\frac{36 - 56}{105}) = \frac{-20}{105} = \frac{-4}{21}\newlineDenominator: 1+(1235)×(815)=1+(96(35×15))1 + (\frac{12}{35}) \times (\frac{8}{15}) = 1 + (\frac{96}{(35\times15)})\newlineDenominator: 1+(96525)=(525+96525)=621525=2071751 + (\frac{96}{525}) = (\frac{525 + 96}{525}) = \frac{621}{525} = \frac{207}{175}
  6. Divide to Find tan⁡(A−B)\tan(A-B): Now divide the numerator by the denominator to find tan⁡(A−B)\tan(A-B).\newlinetan⁡(A−B)=−421/207175\tan(A-B) = \frac{-4}{21} / \frac{207}{175}\newlinetan⁡(A−B)=−421⋅175207\tan(A-B) = \frac{-4}{21} \cdot \frac{175}{207}
  7. Final Simplification: Simplify the expression by canceling out common factors.\newlinetan⁡(A−B)=(−4×175)/(21×207)\tan(A-B) = (-4 \times 175) / (21 \times 207)\newlinetan⁡(A−B)=(−700)/(4347)\tan(A-B) = (-700) / (4347)\newlinetan⁡(A−B)=−100/621\tan(A-B) = -100 / 621

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