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If cos A=(35)/(37) and sin B=(28)/(53) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If cos⁡A=3537 \cos A=\frac{35}{37} and sin⁡B=2853 \sin B=\frac{28}{53} 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=3537 \cos A=\frac{35}{37} and sin⁡B=2853 \sin B=\frac{28}{53} 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 and cos⁡B\cos B: Use the given values to find sin⁡A\sin A and cos⁡B\cos B. Since cos⁡A=3537\cos A = \frac{35}{37}, we can 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 A sin⁡2A=1−(3537)2\sin^2 A = 1 - \left(\frac{35}{37}\right)^2 sin⁡2A=1−12251369\sin^2 A = 1 - \frac{1225}{1369} cos⁡B\cos B00 cos⁡B\cos B11 cos⁡B\cos B22 cos⁡B\cos B33
  2. Find cos⁡B\cos B using sin⁡B\sin B: Similarly, find cos⁡B\cos B using sin⁡B=2853\sin B = \frac{28}{53}.
    cos⁡2B=1−sin⁡2B\cos^2 B = 1 - \sin^2 B
    cos⁡2B=1−(2853)2\cos^2 B = 1 - \left(\frac{28}{53}\right)^2
    cos⁡2B=1−7842809\cos^2 B = 1 - \frac{784}{2809}
    cos⁡2B=28092809−7842809\cos^2 B = \frac{2809}{2809} - \frac{784}{2809}
    cos⁡2B=20252809\cos^2 B = \frac{2025}{2809}
    cos⁡B=20252809\cos B = \sqrt{\frac{2025}{2809}}
    sin⁡B\sin B00
  3. Use angle sum identity for tangent: Use the angle sum identity for tangent to find tan⁡(A+B)\tan(A+B).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}Since tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A} and tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}, we can substitute the values we found.tan⁡A=12/3735/37=1235\tan A = \frac{12/37}{35/37} = \frac{12}{35}tan⁡B=28/5345/53=2845\tan B = \frac{28/53}{45/53} = \frac{28}{45}tan⁡(A+B)=1235+28451−(1235⋅2845)\tan(A+B) = \frac{\frac{12}{35} + \frac{28}{45}}{1 - \left(\frac{12}{35} \cdot \frac{28}{45}\right)}
  4. Simplify expression for tan⁡(A+B)\tan(A+B): Simplify the expression for tan⁡(A+B)\tan(A+B).tan⁡(A+B)=1235+2845\tan(A+B) = \frac{12}{35} + \frac{28}{45} / 1−(1235⋅2845)1 - \left(\frac{12}{35} \cdot \frac{28}{45}\right)To add the fractions, find a common denominator, which is 35⋅4535\cdot45.tan⁡(A+B)=(12⋅45)+(28⋅35)35⋅45−(12⋅28)\tan(A+B) = \frac{(12\cdot45) + (28\cdot35)}{35\cdot45 - (12\cdot28)}tan⁡(A+B)=540+9801575−336\tan(A+B) = \frac{540 + 980}{1575 - 336}tan⁡(A+B)=15201239\tan(A+B) = \frac{1520}{1239}
  5. Check for simplification: Check for any possible simplification of the fraction.\newlineThe numerator and denominator of 15201239\frac{1520}{1239} do not have any common factors other than 11, so the fraction is already in its simplest form.

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