Question

If sinA=3/5 and tanB=1/2, where (pi/2)<A<pi< B<(3pi/2), then: 8 tanA-✓5 sec B=​

Answer 1

Answer:

given A is in second quadrant and B is in third quadrant.

now, sinA = 3/5 [p/h]

b= √ h^2 - p^2

= √ 5^2 - 3^2

= √25-9

= √16

b= 4

similarly,

tanB = 1/2 [p/b]

h= √p^2 + b^2

= √1^2 + 2^2

= √1+4

h = √5

now, 8 tanA - √5 secB

= 8*[p/b] - √5* [h/b]

= 8* (-3/4) - √5*(-√5/2). (-ve sign as tan is -ve in second quadrant and sec is -ve in third quadrant)

= -6 - (-5/2(

= -6 +5/2

= (-12+5)/2

= -7/2