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If tan A=(35)/(12) and cos B=(21)/(29) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If tan⁡A=3512 \tan A=\frac{35}{12} and cos⁡B=2129 \cos B=\frac{21}{29} 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=3512 \tan A=\frac{35}{12} and cos⁡B=2129 \cos B=\frac{21}{29} 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 know tan⁡A=3512\tan A = \frac{35}{12}, but we need to find tan⁡B\tan B using the given cos⁡B=2129\cos B = \frac{21}{29}.
  2. Find tan⁡B\tan B: To find tan⁡B\tan B, we first need to find sin⁡B\sin B. Since cos⁡2B+sin⁡2B=1\cos^2 B + \sin^2 B = 1, we can solve for sin⁡B\sin B. \newlinesin⁡2B=1−cos⁡2B\sin^2 B = 1 - \cos^2 B\newlinesin⁡2B=1−(2129)2\sin^2 B = 1 - (\frac{21}{29})^2\newlinesin⁡2B=1−(441841)\sin^2 B = 1 - (\frac{441}{841})\newlinesin⁡2B=(841−441)/841\sin^2 B = (841 - 441) / 841\newlinesin⁡2B=400841\sin^2 B = \frac{400}{841}\newlinetan⁡B\tan B00\newlinetan⁡B\tan B11, since tan⁡B\tan B22 is in Quadrant I, sin⁡B\sin B is positive.
  3. Calculate tan⁡B\tan B: Now that we have sin⁡B\sin B, we can find tan⁡B\tan B.
    tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}
    tan⁡B=20/2921/29\tan B = \frac{20/29}{21/29}
    tan⁡B=2021\tan B = \frac{20}{21}
  4. Substitute tan⁡A\tan A and tan⁡B\tan B: Substitute tan⁡A\tan A and tan⁡B\tan B into the tan⁡(A+B)\tan(A+B) formula.\newlinetan⁡(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)=3512+20211−(3512⋅2021)\tan(A+B) = \frac{\frac{35}{12} + \frac{20}{21}}{1 - \left(\frac{35}{12} \cdot \frac{20}{21}\right)}
  5. Simplify numerator and denominator: Simplify the numerator and the denominator separately.\newlineFirst, find a common denominator for the numerator.\newlineThe common denominator for 1212 and 2121 is 8484.\newlinetan⁡(A+B)=(35×712×7+20×421×4)/(1−3512×2021)\tan(A+B) = \left(\frac{35 \times 7}{12 \times 7} + \frac{20 \times 4}{21 \times 4}\right) / \left(1 - \frac{35}{12} \times \frac{20}{21}\right)\newlinetan⁡(A+B)=24584+8084/(1−700252)\tan(A+B) = \frac{245}{84} + \frac{80}{84} / \left(1 - \frac{700}{252}\right)
  6. Continue simplifying: Continue simplifying the numerator and the denominator.\newlinetan⁡(A+B)=32584/(1−700252)\tan(A+B) = \frac{325}{84} / \left(1 - \frac{700}{252}\right)\newlineNow, simplify the denominator:\newline1−700252=252252−7002521 - \frac{700}{252} = \frac{252}{252} - \frac{700}{252}\newline1−700252=252−7002521 - \frac{700}{252} = \frac{252 - 700}{252}\newline1−700252=−4482521 - \frac{700}{252} = \frac{-448}{252}\newline1−700252=−56631 - \frac{700}{252} = \frac{-56}{63} (simplifying the fraction by dividing both numerator and denominator by 44)
  7. Divide numerator by denominator: Now, divide the numerator by the denominator.\newlinetan⁡(A+B)=32584/−5663\tan(A+B) = \frac{325}{84} / \frac{-56}{63}\newlineTo divide by a fraction, multiply by its reciprocal.\newlinetan⁡(A+B)=32584×63−56\tan(A+B) = \frac{325}{84} \times \frac{63}{-56}
  8. Simplify multiplication: Simplify the multiplication.\newlinetan⁡(A+B)=325×6384×−56\tan(A+B) = \frac{325 \times 63}{84 \times -56}\newlinetan⁡(A+B)=20475−4704\tan(A+B) = \frac{20475}{-4704}\newlinetan⁡(A+B)=−204754704\tan(A+B) = -\frac{20475}{4704} (keeping the negative sign in the numerator)
  9. Check for further simplification: Check if the fraction can be simplified further.\newlineBoth 2047520475 and 47044704 are divisible by 33.\newlinetan⁡(A+B)=−68251568\tan(A+B) = \frac{-6825}{1568}\newlineThis fraction cannot be simplified further.

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