Question

If 8 tan A = -15 and 25 sin B = -7 and neither A nor B is in the fourth quadrant, then show that sin A cos B + cos A sin B = $\frac{-304}{425}$

Answer 1

we know, tangent be negative in 2nd or 4th quadrant. while sine be negative in 3rd or 4th quadrant.

a/c to question, If 8 tan A = -15 and 25 sin B = -7 and neither A nor B is in the fourth quadrant.
it means, A must be lie in 2nd quadrant and B must be lie in 3rd quadrant.

given, 8tanA = - 15 => tanA = -15/8
then, sinA = 15/17 [ because sine will be positive in 2nd quadrant ]
cosA = -8/17 [ because cosine will be negative in 2nd quadrant ]

again, 25sinB = -7 => sinB = -7/25
cosB = -24/25 [ because cosine will be negative in 3rd quadrant]


now, LHS = sinAcosB + cosAsinB

= 15/17 × -24/25 + (-8/17) × -7/25

= (15 × -24)/17 × 25 + (8 × 7)/17 × 25

= (-408 + 56)/425

= -352/425

Answer 2

HELLO DEAR,



given:-
8tanA = - 15
=> tanA = -15/8 = p/b
h = √(225 + 64) = √(289) = 17

then, sinA = 15/17 [ because sine is positive in 2nd quadrant ]
cosA = -8/17 [ because cosine is negative in 2nd quadrant ]

25sinB = -7
=> sinB = -7/25
cosB = -24/25 [ because cosine is negative in 3rd quadrant]


now,
sinAcosB + cosAsinB

=> 15/17 × -24/25 + (-8/17) × -7/25

=> (15 × -24)/17 × 25 + (8 × 7)/17 × 25

=> (-408 + 56)/425

=> -352/425




I HOPE IT'S HELP YOU DEAR,
THANKS