If tan A=(60)/(11) and sin B=(3)/(5) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

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If  tan A=(60)/(11) and  sin B=(3)/(5) and angles A and B are in Quadrant I, find the value of  tan(A+B). Answer:

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Given Tangent Formula: We know that the tangent of a sum of two angles A and B is given by the formula: tan(A+B) = (tan A + tan B) / (1 - tan A * tan B). We are given tan A = 60/11. We need to find tan B to use this formula.Find Tangent B: To find tan B, we need to use the identity sin² B + cos² B = 1 to find cos B, since tan B = sin B / cos B. We are given sin B = 3/5. Let's find cos B. (sin B)² + (cos B)² = 1 (3/5)² + (cos B)² = 1 9/25 + (cos B)² = 1Use Trigonometric Identity: Subtract 9/25 from both sides to isolate (cos B)². (cos B)² = 1 - 9/25 (cos B)² = 16/25 Since B is in Quadrant I, where cosine is positive, we take the positive square root. cos B = √(16/25) cos B = 4/5Calculate Tangent B: Now we can find tan B using sin B and cos B. tan B = sin B / cos B tan B = (3/5) / (4/5) tan B = 3/4Calculate Tangent A+B: We can now use the values of tan A and tan B to find tan(A+B). tan(A+B) = (tan A + tan B) / (1 - tan A * tan B) tan(A+B) = (60/11 + 3/4) / (1 - (60/11 * 3/4))Find Common Denominator: First, we need to find a common denominator to add 60/11 and 3/4. The common denominator is 44. tan(A+B) = ((60/11)*(4/4) + (3/4)*(11/11)) / (1 - (60/11 * 3/4)) tan(A+B) = (240/44 + 33/44) / (1 - (180/44))Simplify Numerators: Now we add the numerators and simplify the expression. tan(A+B) = (240 + 33) / 44 / (1 - 180/44) tan(A+B) = 273/44 / (1 - 180/44)Simplify Denominator: Next, we simplify the denominator. 1 - 180/44 = (44/44) - (180/44) 1 - 180/44 = -136/44 1 - 180/44 = -34/11 (simplifying by dividing both numerator and denominator by 4)Divide Numerators: Now we can divide the numerators by the denominator. tan(A+B) = (273/44) / (-34/11) To divide by a fraction, we multiply by its reciprocal. tan(A+B) = (273/44) * (11/-34)Final Calculation: Multiply the numerators and the denominators. tan(A+B) = (273 * 11) / (44 * -34) tan(A+B) = 3003 / -1496 tan(A+B) = -3003 / 1496 (it is conventional to write negative sign in the numerator)

If $\tan A=\frac{60}{11}$ and $\sin B=\frac{3}{5}$ and angles A and B are in Quadrant I, find the value of $\tan (A+B)$ . $\newline$ Answer:

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If tan⁡A=6011 \tan A=\frac{60}{11} tanA=1160​ and sin⁡B=35 \sin B=\frac{3}{5} sinB=53​ and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) tan(A+B).\newlineAnswer:

If $\tan A=\frac{60}{11}$ and $\sin B=\frac{3}{5}$ and angles A and B are in Quadrant I, find the value of $\tan (A+B)$ . $\newline$ Answer: