If $\tan A=\frac{40}{9}$ and $\sin B=\frac{15}{17}$ and angles A and B are in Quadrant I, find the value of $\tan (A+B)$ . $\newline$ Answer:

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Find trigonometric ratios using multiple identities

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Q. If

\tan A=\frac{40}{9}

and

\sin B=\frac{15}{17}

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

\tan (A+B)

\newline

Answer:

Use $\tan(A+B)$ formula: Use the formula for $\tan(A+B)$ , which is $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . We know $\tan A = \frac{40}{9}$ . We need to find $\tan B$ using $\sin B = \frac{15}{17}$ .
Find $\tan B$ using $\sin B$ : To find $\tan B$ , we need to find $\cos B$ . Since $\sin^2 B + \cos^2 B = 1$ , we can solve for $\cos B$ .
$\cos B = \sqrt{1 - \sin^2 B}$
$\cos B = \sqrt{1 - (\frac{15}{17})^2}$
$\cos B = \sqrt{1 - \frac{225}{289}}$
$\cos B = \sqrt{\frac{289 - 225}{289}}$
$\sin B$ $0$
$\sin B$ $1$
Find $\cos B$ : Now that we have $\cos B$ , we can find $\tan B$ using the identity $\tan B = \frac{\sin B}{\cos B}$ . $\newline$ $\tan B = \frac{\frac{15}{17}}{\frac{8}{17}}$ $\newline$ $\tan B = \frac{15}{8}$
Find $\tan B$ : Substitute the values of $\tan A$ and $\tan B$ into the $\tan(A+B)$ formula. $\newline$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $\newline$ $\tan(A+B) = \frac{(\frac{40}{9}) + (\frac{15}{8})}{1 - (\frac{40}{9}) \cdot (\frac{15}{8})}$
Substitute values into formula: Calculate the numerator and denominator separately. $\newline$ Numerator: $(\frac{40}{9}) + (\frac{15}{8}) = (\frac{320}{72}) + (\frac{135}{72}) = (\frac{320 + 135}{72}) = \frac{455}{72}$ $\newline$ Denominator: $1 - (\frac{40}{9}) \times (\frac{15}{8}) = 1 - (\frac{600}{72}) = (\frac{72}{72}) - (\frac{600}{72}) = \frac{-528}{72}$
Calculate numerator and denominator: Now, divide the numerator by the denominator to find $\tan(A+B)$ . $\tan(A+B) = \frac{455}{72} / \frac{-528}{72}$ $\tan(A+B) = \frac{455}{-528}$ $\tan(A+B) = -\frac{455}{528}$
Divide numerator by denominator: Simplify the fraction to its lowest terms. $\newline$ $\tan(A+B) = -\frac{455}{528}$ can be simplified by dividing both numerator and denominator by their greatest common divisor, which is $1$ in this case. $\newline$ $\tan(A+B) = -\frac{455}{528}$