Answer:
sin (2A+2B) = 0
Step-by-step explanation:
Given that,
cosec A= -2 and sec B= -2, A and B lie in the same quadrant.
We need to find the value of sin (2A+2B).
[tex]A=cosec^{-1}(-2)\\\\=-30[/tex]
And
[tex]B=\sec^{-1}(-2)\\\\=120^{\circ}[/tex]
To find the value of sin (2A+2B) as follows :
sin [2(-30)+2(120)] = sin (-60+240)
=sin (180)
= 0
So, the value of sin (2A+2B) is qual to 0.
