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If tan A=(28)/(45) and cos B=(4)/(5) and angles A and B are in Quadrant I, find the value of tan(A-B). Answer:

If tan⁡A=2845 \tan A=\frac{28}{45} and cos⁡B=45 \cos B=\frac{4}{5} 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=2845 \tan A=\frac{28}{45} and cos⁡B=45 \cos B=\frac{4}{5} and angles A and B are in Quadrant I, find the value of tan⁡(A−B) \tan (A-B) .\newlineAnswer:
  1. Given tan⁡(A)\tan(A): We know that tan⁡(A)=2845\tan(A) = \frac{28}{45}. To find tan⁡(A−B)\tan(A-B), we can use the formula for the tangent of the difference 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⁡B\tan B, which requires finding sin⁡B\sin B since we already have cos⁡B\cos B.
  2. Find sin⁡B\sin B: We have cos⁡B=45\cos B = \frac{4}{5}. Since BB is in Quadrant I, where all trigonometric functions are positive, we can find sin⁡B\sin B using the Pythagorean identity sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1. Let's calculate sin⁡B\sin B.\newlinesin⁡2B=1−cos⁡2B\sin^2 B = 1 - \cos^2 B\newlinesin⁡2B=1−(45)2\sin^2 B = 1 - \left(\frac{4}{5}\right)^2\newlinesin⁡2B=1−1625\sin^2 B = 1 - \frac{16}{25}\newlinesin⁡2B=2525−1625\sin^2 B = \frac{25}{25} - \frac{16}{25}\newlinecos⁡B=45\cos B = \frac{4}{5}00\newlinecos⁡B=45\cos B = \frac{4}{5}11\newlinecos⁡B=45\cos B = \frac{4}{5}22
  3. Find tan⁡B\tan B: Now that we have sin⁡B\sin B, we can find tan⁡B\tan B using the definition tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}.
    tan⁡B=3/54/5\tan B = \frac{3/5}{4/5}
    tan⁡B=35×54\tan B = \frac{3}{5} \times \frac{5}{4}
    tan⁡B=34\tan B = \frac{3}{4}
  4. Use formula for tan⁡(A−B)\tan(A-B): With tan⁡A=2845\tan A = \frac{28}{45} and tan⁡B=34\tan B = \frac{3}{4}, we can now use the formula for tan⁡(A−B)\tan(A-B).\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)=2845−341+(2845⋅34)\tan(A-B) = \frac{\frac{28}{45} - \frac{3}{4}}{1 + (\frac{28}{45} \cdot \frac{3}{4})}
  5. Simplify numerator: Let's simplify the numerator and the denominator separately.\newlineFirst, the numerator:\newlinetan⁡A−tan⁡B=2845−34\tan A - \tan B = \frac{28}{45} - \frac{3}{4}\newlineTo subtract these fractions, we need a common denominator, which is 180180.\newlinetan⁡A−tan⁡B=28×445×4−3×454×45\tan A - \tan B = \frac{28 \times 4}{45 \times 4} - \frac{3 \times 45}{4 \times 45}\newlinetan⁡A−tan⁡B=112180−135180\tan A - \tan B = \frac{112}{180} - \frac{135}{180}\newlinetan⁡A−tan⁡B=112−135180\tan A - \tan B = \frac{112 - 135}{180}\newlinetan⁡A−tan⁡B=−23180\tan A - \tan B = -\frac{23}{180}
  6. Simplify denominator: Now, let's simplify the denominator:\newline1+tan⁡A⋅tan⁡B=1+(2845⋅34)1 + \tan A \cdot \tan B = 1 + \left(\frac{28}{45} \cdot \frac{3}{4}\right)\newline1+tan⁡A⋅tan⁡B=1+(28⋅345⋅4)1 + \tan A \cdot \tan B = 1 + \left(\frac{28 \cdot 3}{45 \cdot 4}\right)\newline1+tan⁡A⋅tan⁡B=1+841801 + \tan A \cdot \tan B = 1 + \frac{84}{180}\newline1+tan⁡A⋅tan⁡B=1+7151 + \tan A \cdot \tan B = 1 + \frac{7}{15}\newlineTo add 11 to 715\frac{7}{15}, we need to express 11 as 1515\frac{15}{15}.\newline1+tan⁡A⋅tan⁡B=1515+7151 + \tan A \cdot \tan B = \frac{15}{15} + \frac{7}{15}\newline1+tan⁡A⋅tan⁡B=(15+7)151 + \tan A \cdot \tan B = \frac{(15 + 7)}{15}\newline1+tan⁡A⋅tan⁡B=1+(28⋅345⋅4)1 + \tan A \cdot \tan B = 1 + \left(\frac{28 \cdot 3}{45 \cdot 4}\right)00
  7. Divide numerator by denominator: Now we can divide the numerator by the denominator to find tan⁡(A−B)\tan(A-B).tan⁡(A−B)=−23180/2215\tan(A-B) = \frac{-23}{180} / \frac{22}{15}To divide by a fraction, we multiply by its reciprocal.tan⁡(A−B)=−23180×1522\tan(A-B) = \frac{-23}{180} \times \frac{15}{22}
  8. Simplify multiplication: Let's simplify the multiplication by canceling out common factors.\newlinetan⁡(A−B)=(−23×15)/(180×22)\tan(A-B) = (-23 \times 15) / (180 \times 22)\newlinetan⁡(A−B)=(−345)/(3960)\tan(A-B) = (-345) / (3960)\newlineBoth numerator and denominator are divisible by 1515.\newlinetan⁡(A−B)=(−23)/(264)\tan(A-B) = (-23) / (264)
  9. Further simplify: We can further simplify by dividing both numerator and denominator by 2323. \newlinetan⁡(A−B)=−1264/23\tan(A-B) = \frac{-1}{264/23}\newlinetan⁡(A−B)=−111.4782608696\tan(A-B) = \frac{-1}{11.4782608696}

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