If $\sin A=\frac{35}{37}$ and $\cos B=\frac{8}{17}$ 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

\sin A=\frac{35}{37}

and

\cos B=\frac{8}{17}

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

\tan (A+B)

\newline

Answer:

Find Tangent Values: Use the sine and cosine values to find the corresponding tangent values for angles $A$ and $B$ using the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ . For angle $A$ , we have $\sin A = \frac{35}{37}$ . To find $\cos A$ , we use the Pythagorean identity $\sin^2(A) + \cos^2(A) = 1$ . $\cos^2(A) = 1 - \sin^2(A)$ $\cos^2(A) = 1 - \left(\frac{35}{37}\right)^2$ $\cos^2(A) = 1 - \frac{1225}{1369}$ $B$ $0$ $B$ $1$ $B$ $2$ $B$ $3$ Now, $B$ $4$ .
Calculate Angle A: For angle B, we have $\cos B = \frac{8}{17}$ . To find $\sin B$ , we use the Pythagorean identity $\sin^2(B) + \cos^2(B) = 1$ . $\newline$ $\sin^2(B) = 1 - \cos^2(B)$ $\newline$ $\sin^2(B) = 1 - \left(\frac{8}{17}\right)^2$ $\newline$ $\sin^2(B) = 1 - \frac{64}{289}$ $\newline$ $\sin^2(B) = \frac{289 - 64}{289}$ $\newline$ $\sin^2(B) = \frac{225}{289}$ $\newline$ $\sin(B) = \sqrt{\frac{225}{289}}$ $\newline$ $\sin(B) = \frac{15}{17}$ $\newline$ Now, $\sin B$ $0$ .
Calculate Angle B: Use the angle sum identity for tangent to find $\tan(A+B)$ : $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ Substitute the values of $\tan A$ and $\tan B$ into the formula: $\tan(A+B) = \frac{\frac{35}{12} + \frac{15}{8}}{1 - (\frac{35}{12} \cdot \frac{15}{8})}$
Use Angle Sum Identity: Simplify the numerator and the denominator separately: $\newline$ $\tan(A+B) = \frac{(\frac{35}{12}) \cdot (\frac{2}{2}) + (\frac{15}{8}) \cdot (\frac{3}{3})}{1 - (\frac{35}{12} \cdot \frac{15}{8})}$ $\newline$ $\tan(A+B) = \frac{\frac{70}{24} + \frac{45}{24}}{1 - \frac{525}{96}}$ $\newline$ $\tan(A+B) = \frac{\frac{115}{24}}{1 - \frac{525}{96}}$
Simplify Numerator and Denominator: Simplify the denominator further: $\newline$ $\tan(A+B) = \frac{115}{24} / \left(\frac{96}{96} - \frac{525}{96}\right)$ $\newline$ $\tan(A+B) = \frac{115}{24} / \left(\frac{96 - 525}{96}\right)$ $\newline$ $\tan(A+B) = \frac{115}{24} / \left(-\frac{429}{96}\right)$
Simplify Denominator: Simplify the entire expression by multiplying the numerator by the reciprocal of the denominator: $\newline$ $\tan(A+B) = \frac{115}{24} \times \frac{96}{-429}$ $\newline$ Simplify the fraction by canceling common factors: $\newline$ $\tan(A+B) = \frac{115 \times 4}{-429}$ $\newline$ $\tan(A+B) = \frac{460}{-429}$
Final Simplification: Reduce the fraction to its simplest form: $\tan(A+B) = -\frac{460}{429}$