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If cos A=(21)/(29) 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=2129 \cos A=\frac{21}{29} 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=2129 \cos A=\frac{21}{29} 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. Find tan⁡A\tan A: Use the identity for the tangent of the sum of two angles: 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}. First, we need to find tan⁡A\tan A. Since we know cos⁡A=2129\cos A = \frac{21}{29} and AA is in Quadrant I, we can find sin⁡A\sin A using the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A sin⁡2A=1−(2129)2\sin^2 A = 1 - \left(\frac{21}{29}\right)^2 sin⁡2A=1−441841\sin^2 A = 1 - \frac{441}{841} 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}00 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}11 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}22 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}33 Now, 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}44 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}55 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}66
  2. Substitute values into identity: Now we have tan⁡A=2021\tan A = \frac{20}{21} and tan⁡B=512\tan B = \frac{5}{12}. We can substitute these values into the identity for tan⁡(A+B)\tan(A+B). \newlinetan⁡(A+B)=(tan⁡A+tan⁡B)(1−tan⁡A⋅tan⁡B)\tan(A+B) = \frac{(\tan A + \tan B)}{(1 - \tan A \cdot \tan B)}\newlinetan⁡(A+B)=(2021+512)(1−(2021⋅512))\tan(A+B) = \frac{(\frac{20}{21} + \frac{5}{12})}{(1 - (\frac{20}{21} \cdot \frac{5}{12}))}
  3. Simplify numerator and denominator: Simplify the numerator and the denominator separately.\newlineFor the numerator:\newlinetan⁡A+tan⁡B=2021+512\tan A + \tan B = \frac{20}{21} + \frac{5}{12}\newlineTo add these fractions, find a common denominator, which is 21×12=25221 \times 12 = 252.\newline(2021)×(1212)+(512)×(2121)=240252+105252\left(\frac{20}{21}\right) \times \left(\frac{12}{12}\right) + \left(\frac{5}{12}\right) \times \left(\frac{21}{21}\right) = \frac{240}{252} + \frac{105}{252}\newline=240+105252= \frac{240 + 105}{252}\newline=345252= \frac{345}{252}\newlineFor the denominator:\newline1−(2021×512)=1−1002521 - \left(\frac{20}{21} \times \frac{5}{12}\right) = 1 - \frac{100}{252}\newlineTo subtract these fractions, express 11 as 252252\frac{252}{252}.\newline(252252)−100252=252−100252\left(\frac{252}{252}\right) - \frac{100}{252} = \frac{252 - 100}{252}\newline=152252= \frac{152}{252}
  4. Divide to find tan⁡(A+B)\tan(A+B): Now we can divide the numerator by the denominator to find tan⁡(A+B)\tan(A+B).tan⁡(A+B)=345252/152252\tan(A+B) = \frac{345}{252} / \frac{152}{252}When dividing by a fraction, multiply by the reciprocal of the divisor.tan⁡(A+B)=345252×252152\tan(A+B) = \frac{345}{252} \times \frac{252}{152}The 252252 in the numerator and denominator cancel out.tan⁡(A+B)=345152\tan(A+B) = \frac{345}{152}
  5. Simplify final fraction: Simplify the fraction 345152\frac{345}{152} if possible.\newline345345 and 152152 have no common factors other than 11, so the fraction is already in its simplest form.

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