Math, asked by saniathecutie15, 9 days ago

if 8tanA=-15 and 25 sinB= -7 and neither A and nor B is in 4quadrant then show that sinA cosB+cosAsinB=-304/425​

Answers

Answered by MysticSohamS
2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

to \: prove \: that :  \\ sin \: A.co s\: B + cos \: A.sin \: B =   - \frac{  304}{425}  \\  \\ given :  \\ tan \: A =  -  \frac{8}{15}  \\  \\ sin \: B =  -  \frac{7}{25}  \\  \\ since \: we \: know \: that \\ tangent \: angle \: angle \: is \: always \\ negative \: in \: quadrant \: 2 \: and \: 4 \\ but \: as \: angle \: A \: doesnt \: lie  \: in\\ quadrant \: 4 \:  \\ angle \: A \: must \: lie \: in \: quadrant \: 2 \\  \\ similarly \: we \: know \: that \\ sine \: angle \: is \: always \: negative \: in \\ quadrant \: 3 \: and \:quadrant \:  4 \\ but \: since \: angle \: B \: is \: not \: lying \\ in \: quadrant \: 4 \\ angle \: B \: must \: lie \: in \: quadrant \: 3

so \: now \: using \\ 1 + tan {}^{2}  \: A = sec {}^{2}  \: A \\ \\   = 1 + ( -  \frac{15}{8}  \: ) {}^{2}  \\  \\  = 1 +  \frac{225}{64}  \\  \\  =  \frac{225 + 64}{64}  \\  \\  =  \frac{289}{64}  \\  \\ sec \: A =  \frac{17}{8}  \\  \\ cos \: A =  \frac{8}{17}  \\  \\ since \: angle \:A  \:lies \: in \: quadrant \: 2 \\ cos \: A =   - \frac{8}{17}  \\  \\ similarly \: by \: applying \\ sin {}^{2} \:  A + cos {}^{2}  \: A = 1 \\ we \: get \\  \\ sin \: A =  \frac{15}{17}  \\  \\ since \: angle \: A\: lies \: in \: quadrant \: A \\ sin \: A =  \frac{15}{17}

accordingly \: here \\ sin \: B =  -  \frac{7}{25}  \\  \\ using \: now \\ sin   {}^{2}  \: B + cos {}^{2}  \: B = 1 \\ cos {}^{2}  \: B = 1 - sin {}^{2}  \: B \\  \\  = 1 - ( -  \frac{7}{25}  \: ) {}^{2}  \\  \\  = 1 -  \frac{49}{625}  \\  \\  =  \frac{625 - 49}{625}  \\  \\  =  \frac{576}{625}  \\  \\ cos \: B =  \frac{24}{25}  \\  \\ but \: as \: angle \: B \: lies \: in \: quadrant \: 3 \\ cos \: B =  -  \frac{24}{25}

now \: considering \:  \\ sin \: A. cos \: B+cos \: .Asin \: B=- \frac{304}{425}  \\  \\ LHS = sin \: A .cos \: B+cos \: A.sin \: B \\  \\  =   \frac{15}{17}  \:   \times ( -  \frac{24}{25}  \: ) + ( -  \frac{8}{17} ) \times ( -  \frac{7}{25} ) \\  \\  =  \frac{15( - 24) +( - 7)( - 8) }{17 \times 25}  \\  \\  =  \frac{ - 360 + 56}{425}  \\  \\  =  -  \frac{304}{425}  \\  \\  = RHS

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