Full solution

Q. If

\sin A=\frac{45}{53}

and

\cos B=\frac{12}{13}

and angles A and B are in Quadrant I, find the value of

\tan (A+B)

\newline

Answer:

Find Trigonometric Functions: Use the Pythagorean identity to find $\cos A$ and $\sin B$ . Since $A$ is in Quadrant I, $\cos A$ will be positive. Similarly, since $B$ is in Quadrant I, $\sin B$ will also be positive. For angle $A$ , we have $\sin A = \frac{45}{53}$ . Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ , we can find $\cos A$ . $\sin B$ $0$ . For angle $B$ , we have $\sin B$ $2$ . Using the Pythagorean identity $\sin B$ $3$ , we can find $\sin B$ . $\sin B$ $5$ .
Calculate Tangent of Sum: Use the angle sum identity for tangent to find $\tan(A+B)$ . The angle sum identity for tangent is $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . We need to find $\tan A$ and $\tan B$ . $\tan A = \frac{\sin A}{\cos A} = \frac{45/53}{28/53} = \frac{45}{28}$ . $\tan B = \frac{\sin B}{\cos B} = \frac{5/13}{12/13} = \frac{5}{12}$ .
Substitute and Simplify: Substitute the values of $\tan A$ and $\tan B$ into the angle sum identity. $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{\frac{45}{28} + \frac{5}{12}}{1 - \left(\frac{45}{28} \cdot \frac{5}{12}\right)}$ .
Substitute and Simplify: Substitute the values of $\tan A$ and $\tan B$ into the angle sum identity. $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{45/28 + 5/12}{1 - (45/28 \cdot 5/12)}$ .Perform the addition and multiplication in the numerator and denominator. $\tan(A+B) = \frac{(45/28) \cdot (12/12) + (5/12) \cdot (28/28)}{1 - (45/28) \cdot (5/12)}$ $\tan(A+B) = \frac{(540/336) + (140/336)}{1 - (225/336)}$ $\tan(A+B) = \frac{680/336}{1 - (225/336)}$ $\tan(A+B) = \frac{680/336}{(336/336) - (225/336)}$ $\tan(A+B) = \frac{680/336}{111/336}$
Substitute and Simplify: Substitute the values of $\tan A$ and $\tan B$ into the angle sum identity. $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{45/28 + 5/12}{1 - (45/28 \cdot 5/12)}$ .Perform the addition and multiplication in the numerator and denominator. $\tan(A+B) = \frac{(45/28) \cdot (12/12) + (5/12) \cdot (28/28)}{1 - (45/28) \cdot (5/12)}$ $\tan(A+B) = \frac{(540/336) + (140/336)}{1 - (225/336)}$ $\tan(A+B) = \frac{680/336}{1 - (225/336)}$ $\tan(A+B) = \frac{680/336}{(336/336) - (225/336)}$ $\tan(A+B) = \frac{680/336}{111/336}$ Simplify the fraction. $\tan(A+B) = \frac{680/336} \cdot \frac{336/111} = \frac{680}{111}$