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If cos A=(7)/(25) and tan B=(3)/(4) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If cos⁡A=725 \cos A=\frac{7}{25} and tan⁡B=34 \tan B=\frac{3}{4} 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=725 \cos A=\frac{7}{25} and tan⁡B=34 \tan B=\frac{3}{4} 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=725\cos A = \frac{7}{25} 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 Asin⁡2A=1−(725)2\sin^2 A = 1 - \left(\frac{7}{25}\right)^2sin⁡2A=1−49625\sin^2 A = 1 - \frac{49}{625}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}00tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}11tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}22tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}33Now, 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.
  2. Substitute values into identity: Now we have tan⁡A=247\tan A = \frac{24}{7} and tan⁡B=34\tan B = \frac{3}{4}. We can substitute these values into the identity for tan⁡(A+B)\tan(A+B).tan⁡(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)}tan⁡(A+B)=(247+34)(1−(247⋅34))\tan(A+B) = \frac{(\frac{24}{7} + \frac{3}{4})}{(1 - (\frac{24}{7} \cdot \frac{3}{4}))}
  3. Simplify numerator and denominator: Simplify the numerator and the denominator separately.\newlineFor the numerator:\newlinetan⁡A+tan⁡B=247+34\tan A + \tan B = \frac{24}{7} + \frac{3}{4}\newlineTo add these fractions, find a common denominator, which is 2828.\newlinetan⁡A+tan⁡B=24×47×4+3×74×7\tan A + \tan B = \frac{24 \times 4}{7 \times 4} + \frac{3 \times 7}{4 \times 7}\newlinetan⁡A+tan⁡B=9628+2128\tan A + \tan B = \frac{96}{28} + \frac{21}{28}\newlinetan⁡A+tan⁡B=96+2128\tan A + \tan B = \frac{96 + 21}{28}\newlinetan⁡A+tan⁡B=11728\tan A + \tan B = \frac{117}{28}\newlineFor the denominator:\newline1−tan⁡A×tan⁡B=1−(247×34)1 - \tan A \times \tan B = 1 - \left(\frac{24}{7} \times \frac{3}{4}\right)\newline1−tan⁡A×tan⁡B=1−(7228)1 - \tan A \times \tan B = 1 - \left(\frac{72}{28}\right)\newline1−tan⁡A×tan⁡B=2828−72281 - \tan A \times \tan B = \frac{28}{28} - \frac{72}{28}\newline1−tan⁡A×tan⁡B=−44281 - \tan A \times \tan B = \frac{-44}{28}
  4. Divide to find tan⁡(A+B)\tan(A+B): Now, divide the numerator by the denominator to find tan⁡(A+B)\tan(A+B).tan⁡(A+B)=11728/(−4428)\tan(A+B) = \frac{117}{28} / \left(-\frac{44}{28}\right)When dividing by a fraction, multiply by the reciprocal of the denominator.tan⁡(A+B)=11728×(28−44)\tan(A+B) = \frac{117}{28} \times \left(\frac{28}{-44}\right)The 2828s cancel out.tan⁡(A+B)=117−44\tan(A+B) = \frac{117}{-44}Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 11 in this case.tan⁡(A+B)=−11744\tan(A+B) = -\frac{117}{44}

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