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If tan A=(40)/(9) and sin B=(15)/(17) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If tan⁡A=409 \tan A=\frac{40}{9} and sin⁡B=1517 \sin B=\frac{15}{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 tan⁡A=409 \tan A=\frac{40}{9} and sin⁡B=1517 \sin B=\frac{15}{17} 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) formula: Use the formula 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 know tan⁡A=409\tan A = \frac{40}{9}. We need to find tan⁡B\tan B using sin⁡B=1517\sin B = \frac{15}{17}.
  2. Find tan⁡B\tan B using sin⁡B\sin B: To find tan⁡B\tan B, we need to find cos⁡B\cos B. Since sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1, we can solve for cos⁡B\cos B.
    cos⁡B=1−sin⁡2B\cos B = \sqrt{1 - \sin^2 B}
    cos⁡B=1−(1517)2\cos B = \sqrt{1 - (\frac{15}{17})^2}
    cos⁡B=1−225289\cos B = \sqrt{1 - \frac{225}{289}}
    cos⁡B=289−225289\cos B = \sqrt{\frac{289 - 225}{289}}
    sin⁡B\sin B00
    sin⁡B\sin B11
  3. Find cos⁡B\cos B: Now that we have cos⁡B\cos B, we can find tan⁡B\tan B using the identity tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}.\newlinetan⁡B=1517817\tan B = \frac{\frac{15}{17}}{\frac{8}{17}}\newlinetan⁡B=158\tan B = \frac{15}{8}
  4. Find tan⁡B\tan B: Substitute the values of tan⁡A\tan A and tan⁡B\tan B into the tan⁡(A+B)\tan(A+B) formula.\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)+(158)1−(409)⋅(158)\tan(A+B) = \frac{(\frac{40}{9}) + (\frac{15}{8})}{1 - (\frac{40}{9}) \cdot (\frac{15}{8})}
  5. Substitute values into formula: Calculate the numerator and denominator separately.\newlineNumerator: (409)+(158)=(32072)+(13572)=(320+13572)=45572(\frac{40}{9}) + (\frac{15}{8}) = (\frac{320}{72}) + (\frac{135}{72}) = (\frac{320 + 135}{72}) = \frac{455}{72}\newlineDenominator: 1−(409)×(158)=1−(60072)=(7272)−(60072)=−528721 - (\frac{40}{9}) \times (\frac{15}{8}) = 1 - (\frac{600}{72}) = (\frac{72}{72}) - (\frac{600}{72}) = \frac{-528}{72}
  6. Calculate numerator and denominator: Now, divide the numerator by the denominator to find tan⁡(A+B)\tan(A+B).tan⁡(A+B)=45572/−52872\tan(A+B) = \frac{455}{72} / \frac{-528}{72}tan⁡(A+B)=455−528\tan(A+B) = \frac{455}{-528}tan⁡(A+B)=−455528\tan(A+B) = -\frac{455}{528}
  7. Divide numerator by denominator: Simplify the fraction to its lowest terms.\newlinetan⁡(A+B)=−455528\tan(A+B) = -\frac{455}{528} can be simplified by dividing both numerator and denominator by their greatest common divisor, which is 11 in this case.\newlinetan⁡(A+B)=−455528\tan(A+B) = -\frac{455}{528}

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