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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Q. 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:

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