If tanθ = 15/8 and cosθ is negative, find the value of:

[sin(-θ)-cosθ] / [tan(-θ)+sec(-θ)]

AB = 15, BC = 8

Since, tan θ is (+)ve and tan is always (+)ve in the III quadrant, so value of θ lies in the third quadrant.

(In 3rd quadrant, value of sin θ, cos θ and sec θ is (–)ve.)


  • 0
What are you looking for?