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If cos A=(9)/(41) 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=941 \cos A=\frac{9}{41} 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=941 \cos A=\frac{9}{41} 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. Use tan⁡(A−B)\tan(A-B) Identity: Use the identity 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 need to find tan⁡A\tan A and tan⁡B\tan B to use this formula.
  2. Find sin⁡A\sin A: We know that cos⁡A=941\cos A = \frac{9}{41} and tan⁡B=512\tan B = \frac{5}{12}. Since AA is in Quadrant I, sin⁡A\sin A will be positive. Use the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 to find sin⁡A\sin A.sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 Asin⁡2A=1−(941)2\sin^2 A = 1 - \left(\frac{9}{41}\right)^2sin⁡2A=1−811681\sin^2 A = 1 - \frac{81}{1681}cos⁡A=941\cos A = \frac{9}{41}00cos⁡A=941\cos A = \frac{9}{41}11cos⁡A=941\cos A = \frac{9}{41}22cos⁡A=941\cos A = \frac{9}{41}33
  3. Find tan⁡A\tan A: Now, find tan⁡A\tan A using the definition tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}.tan⁡A=(40/41)(9/41)\tan A = \frac{(40/41)}{(9/41)}tan⁡A=409\tan A = \frac{40}{9}
  4. Substitute tan⁡A\tan A and tan⁡B\tan B: Substitute the values of tan⁡A\tan A and tan⁡B\tan B into the identity 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)=409−5121+(409)⋅(512)\tan(A-B) = \frac{\frac{40}{9} - \frac{5}{12}}{1 + \left(\frac{40}{9}\right) \cdot \left(\frac{5}{12}\right)}
  5. Simplify with Common Denominator: Simplify the expression by finding a common denominator for tan⁡A\tan A and tan⁡B\tan B.tan⁡(A−B)=(409×1212−512×99)/(1+409×512)\tan(A-B) = \left(\frac{40}{9} \times \frac{12}{12} - \frac{5}{12} \times \frac{9}{9}\right) / \left(1 + \frac{40}{9} \times \frac{5}{12}\right)tan⁡(A−B)=(480108−45108)/(1+200108)\tan(A-B) = \left(\frac{480}{108} - \frac{45}{108}\right) / \left(1 + \frac{200}{108}\right)tan⁡(A−B)=(435108)/(1+200108)\tan(A-B) = \left(\frac{435}{108}\right) / \left(1 + \frac{200}{108}\right)
  6. Simplify Denominator: Simplify the denominator of the expression. \newline1+200108=108108+2001081 + \frac{200}{108} = \frac{108}{108} + \frac{200}{108}\newline1+200108=3081081 + \frac{200}{108} = \frac{308}{108}
  7. Divide Numerator by Denominator: Now, divide the numerator by the denominator.\newlinetan⁡(A−B)=435108/308108\tan(A-B) = \frac{435}{108} / \frac{308}{108}\newlinetan⁡(A−B)=435108×108308\tan(A-B) = \frac{435}{108} \times \frac{108}{308}
  8. Simplify Expression: Simplify the expression by canceling out the common factor of 108108. \newlinetan⁡(A−B)=435308\tan(A-B) = \frac{435}{308}
  9. Check for Further Simplification: Check if the fraction can be simplified further. 435435 and 308308 have no common factors other than 11, so the fraction is already in its simplest form.

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