`pi lt A lt (3pi)/(2) rArr A` lies in 3rd quadrant,
`sinA=-4/5`
`rArr "cosec"A=1/(sinA)=-5/4`
`cot^(2)A="cosec"^(2)A-1`
`=(-5/4)^(2)-1=9/16`
`cotA=3/4` (In 3rd quadrant, cot is positive)
`therefore tan A=1/(cotA)=4/3`
`secA=(sinA)/(cosA). 1/(sinA)`
`=tanA. "cosec"A`
`4/3(-5/4)=-5/3`
Now `("cosec"A+cotA)/(secA-tanA)=(-5//4)=-5/3`
Now `("cosec"A+cotA)/(secA-tanA)=(-5/4+3/4)/(-5/3-4/3)=(-2/4)/(-9/3)=1/6` Ans
