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If cos A=(12)/(37) and sin B=(3)/(5) and angles A and B are in Quadrant I, find the value of tan(A-B). Answer:

If cos⁡A=1237 \cos A=\frac{12}{37} and sin⁡B=35 \sin B=\frac{3}{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 cos⁡A=1237 \cos A=\frac{12}{37} and sin⁡B=35 \sin B=\frac{3}{5} and angles A and B are in Quadrant I, find the value of tan⁡(A−B) \tan (A-B) .\newlineAnswer:
  1. Find sin⁡A\sin A: We know that cos⁡A=1237\cos A = \frac{12}{37}. Since AA is in Quadrant I, where both sine and cosine are positive, we can find sin⁡A\sin A using the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. First, calculate sin⁡2A\sin^2 A: sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A sin⁡2A=1−(1237)2\sin^2 A = 1 - \left(\frac{12}{37}\right)^2 sin⁡2A=1−1441369\sin^2 A = 1 - \frac{144}{1369} sin⁡2A=13691369−1441369\sin^2 A = \frac{1369}{1369} - \frac{144}{1369} cos⁡A=1237\cos A = \frac{12}{37}00 Now, take the square root to find sin⁡A\sin A: cos⁡A=1237\cos A = \frac{12}{37}22 cos⁡A=1237\cos A = \frac{12}{37}33
  2. Find cos⁡B\cos B: We also know that sin⁡B=35\sin B = \frac{3}{5}. Since B is in Quadrant I, where both sine and cosine are positive, we can find cos⁡B\cos B using the Pythagorean identity sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1.\newlineFirst, calculate cos⁡2B\cos^2 B:\newlinecos⁡2B=1−sin⁡2B\cos^2 B = 1 - \sin^2 B\newlinecos⁡2B=1−(35)2\cos^2 B = 1 - \left(\frac{3}{5}\right)^2\newlinecos⁡2B=1−925\cos^2 B = 1 - \frac{9}{25}\newlinecos⁡2B=2525−925\cos^2 B = \frac{25}{25} - \frac{9}{25}\newlinecos⁡2B=1625\cos^2 B = \frac{16}{25}\newlineNow, take the square root to find cos⁡B\cos B:\newlinesin⁡B=35\sin B = \frac{3}{5}11\newlinesin⁡B=35\sin B = \frac{3}{5}22
  3. Calculate tan⁡(A−B)\tan(A-B): Now we need to find tan⁡(A−B)\tan(A-B). We can use the formula for the tangent of the difference of two angles:\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}\newlineWe already have sin⁡A\sin A and cos⁡A\cos A, and sin⁡B\sin B and cos⁡B\cos B, so we can find tan⁡A\tan A and tan⁡B\tan B:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}\newlinetan⁡A=35/3712/37\tan A = \frac{35/37}{12/37}\newlinetan⁡A=3512\tan A = \frac{35}{12}\newlinetan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}\newlinetan⁡B=3/54/5\tan B = \frac{3/5}{4/5}\newlinetan⁡B=34\tan B = \frac{3}{4}
  4. Calculate tan⁡(A−B)\tan(A-B): Now we need to find tan⁡(A−B)\tan(A-B). We can use the formula for the tangent of the difference of two angles:\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}\newlineWe already have sin⁡A\sin A and cos⁡A\cos A, and sin⁡B\sin B and cos⁡B\cos B, so we can find tan⁡A\tan A and tan⁡B\tan B:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}\newlinetan⁡A=35/3712/37\tan A = \frac{35/37}{12/37}\newlinetan⁡A=3512\tan A = \frac{35}{12}\newlinetan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}\newlinetan⁡B=3/54/5\tan B = \frac{3/5}{4/5}\newlinetan⁡B=34\tan B = \frac{3}{4}Substitute tan⁡A\tan A and tan⁡B\tan B into 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)=3512−341+(3512)⋅(34)\tan(A-B) = \frac{\frac{35}{12} - \frac{3}{4}}{1 + \left(\frac{35}{12}\right) \cdot \left(\frac{3}{4}\right)}\newlineTo subtract the fractions, find a common denominator:\newlinetan⁡(A−B)=3512−9121+(3512)⋅(34)\tan(A-B) = \frac{\frac{35}{12} - \frac{9}{12}}{1 + \left(\frac{35}{12}\right) \cdot \left(\frac{3}{4}\right)}\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}00\newlineNow simplify the fractions:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}11\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}22\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}33\newlineNow multiply by the reciprocal of the denominator:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}44\newlineSimplify the fraction by canceling common factors:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}55\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}66\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}77\newlineFinally, simplify the fraction:\newlinetan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}88

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