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If sin A=(35)/(37) and cos B=(8)/(17) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If sin⁡A=3537 \sin A=\frac{35}{37} and cos⁡B=817 \cos B=\frac{8}{17} 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 sin⁡A=3537 \sin A=\frac{35}{37} and cos⁡B=817 \cos B=\frac{8}{17} and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer:
  1. Find Tangent Values: Use the sine and cosine values to find the corresponding tangent values for angles AA and BB using the identity tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}. For angle AA, we have sin⁡A=3537\sin A = \frac{35}{37}. To find cos⁡A\cos A, we use the Pythagorean identity sin⁡2(A)+cos⁡2(A)=1\sin^2(A) + \cos^2(A) = 1. cos⁡2(A)=1−sin⁡2(A)\cos^2(A) = 1 - \sin^2(A) cos⁡2(A)=1−(3537)2\cos^2(A) = 1 - \left(\frac{35}{37}\right)^2 cos⁡2(A)=1−12251369\cos^2(A) = 1 - \frac{1225}{1369} BB00 BB11 BB22 BB33 Now, BB44.
  2. Calculate Angle A: For angle B, we have cos⁡B=817\cos B = \frac{8}{17}. To find sin⁡B\sin B, we use the Pythagorean identity sin⁡2(B)+cos⁡2(B)=1\sin^2(B) + \cos^2(B) = 1. \newlinesin⁡2(B)=1−cos⁡2(B)\sin^2(B) = 1 - \cos^2(B)\newlinesin⁡2(B)=1−(817)2\sin^2(B) = 1 - \left(\frac{8}{17}\right)^2\newlinesin⁡2(B)=1−64289\sin^2(B) = 1 - \frac{64}{289}\newlinesin⁡2(B)=289−64289\sin^2(B) = \frac{289 - 64}{289}\newlinesin⁡2(B)=225289\sin^2(B) = \frac{225}{289}\newlinesin⁡(B)=225289\sin(B) = \sqrt{\frac{225}{289}}\newlinesin⁡(B)=1517\sin(B) = \frac{15}{17}\newlineNow, sin⁡B\sin B00.
  3. Calculate Angle B: 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}Substitute the values of tan⁡A\tan A and tan⁡B\tan B into the formula:tan⁡(A+B)=3512+1581−(3512⋅158)\tan(A+B) = \frac{\frac{35}{12} + \frac{15}{8}}{1 - (\frac{35}{12} \cdot \frac{15}{8})}
  4. Use Angle Sum Identity: Simplify the numerator and the denominator separately:\newlinetan⁡(A+B)=(3512)⋅(22)+(158)⋅(33)1−(3512⋅158)\tan(A+B) = \frac{(\frac{35}{12}) \cdot (\frac{2}{2}) + (\frac{15}{8}) \cdot (\frac{3}{3})}{1 - (\frac{35}{12} \cdot \frac{15}{8})}\newlinetan⁡(A+B)=7024+45241−52596\tan(A+B) = \frac{\frac{70}{24} + \frac{45}{24}}{1 - \frac{525}{96}}\newlinetan⁡(A+B)=115241−52596\tan(A+B) = \frac{\frac{115}{24}}{1 - \frac{525}{96}}
  5. Simplify Numerator and Denominator: Simplify the denominator further: \newlinetan⁡(A+B)=11524/(9696−52596)\tan(A+B) = \frac{115}{24} / \left(\frac{96}{96} - \frac{525}{96}\right)\newlinetan⁡(A+B)=11524/(96−52596)\tan(A+B) = \frac{115}{24} / \left(\frac{96 - 525}{96}\right)\newlinetan⁡(A+B)=11524/(−42996)\tan(A+B) = \frac{115}{24} / \left(-\frac{429}{96}\right)
  6. Simplify Denominator: Simplify the entire expression by multiplying the numerator by the reciprocal of the denominator:\newlinetan⁡(A+B)=11524×96−429\tan(A+B) = \frac{115}{24} \times \frac{96}{-429}\newlineSimplify the fraction by canceling common factors:\newlinetan⁡(A+B)=115×4−429\tan(A+B) = \frac{115 \times 4}{-429}\newlinetan⁡(A+B)=460−429\tan(A+B) = \frac{460}{-429}
  7. Final Simplification: Reduce the fraction to its simplest form: tan⁡(A+B)=−460429\tan(A+B) = -\frac{460}{429}

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