If $\cos A=\frac{4}{5}$ and $\tan B=\frac{11}{60}$ and angles A and B are in Quadrant I, find the value of $\tan (A+B)$ . $\newline$ Answer:

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Q. If

\cos A=\frac{4}{5}

and

\tan B=\frac{11}{60}

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

\tan (A+B)

\newline

Answer:

Find $\sin A$ : We know that $\cos A = \frac{4}{5}$ . To find $\tan A$ , we need to find $\sin A$ using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ . $\newline$ Substitute $\cos A = \frac{4}{5}$ into the identity: $\newline$ (\sin A)^\(2 + \left(\frac{ $4$ }{ $5$ }\right)^ $2$ = $1$
Simplify equation: Simplify the equation to find $(\sin A)^2$ : $(\sin A)^2 + \frac{16}{25} = 1$ $(\sin A)^2 = 1 - \frac{16}{25}$ $(\sin A)^2 = \frac{25}{25} - \frac{16}{25}$ $(\sin A)^2 = \frac{9}{25}$
Take square root: Take the square root of both sides to find $\sin A$ : $\sin A = \pm\sqrt{\frac{9}{25}}$ Since angle A is in Quadrant I, where sine is positive, we choose the positive root: $\sin A = \frac{3}{5}$
Find $\tan A$ : Now we can find $\tan A$ using $\sin A$ and $\cos A$ :
$\tan A = \frac{\sin A}{\cos A}$
$\tan A = \frac{3/5}{4/5}$
$\tan A = \frac{3}{4}$
Use angle sum identity: We are given $\tan B = \frac{11}{60}$ . To find $\tan(A+B)$ , we use the angle sum identity for tangent: $\newline$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $\newline$ Substitute $\tan A = \frac{3}{4}$ and $\tan B = \frac{11}{60}$ into the identity: $\newline$ $\tan(A+B) = \frac{\frac{3}{4} + \frac{11}{60}}{1 - \left(\frac{3}{4}\right) \cdot \left(\frac{11}{60}\right)}$
Find common denominator: Find a common denominator and simplify the numerator and denominator: $\newline$ $\tan(A+B) = \frac{45/60 + 11/60}{1 - 33/240}$ $\newline$ $\tan(A+B) = \frac{56/60}{1 - 33/240}$ $\newline$ $\tan(A+B) = \frac{56/60}{(240/240) - (33/240)}$ $\newline$ $\tan(A+B) = \frac{56/60}{207/240}$
Simplify fractions: Simplify the fractions and perform the division: $\newline$ $\tan(A+B) = \frac{14}{15} / \frac{207}{240}$ $\newline$ $\tan(A+B) = \frac{14}{15} \times \frac{240}{207}$ $\newline$ $\tan(A+B) = \frac{14 \times 240}{15 \times 207}$
Perform division: Simplify the multiplication and division: $\tan(A+B) = \frac{3360}{3105}$ Now we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $15$ : $\tan(A+B) = \frac{224}{207}$