We have to break each degree in terms of 90
A) 
Which is in third quadrant, therefore sine is negative hence

B) 
Which is in third quadrant, therefore cosine is negative hence

C) 
Which is in third quadrant, therefore tangent is positive hence

D) 
Which is in third quadrant, therefore cosec is negative hence
not defined
E)
Which is in third quadrant, therefore secant is negative hence

F) 
Which is in third quadrant, therefore tangent is positive hence
not defined
Hence only
and
have value -1
Hope this will help
Answer:
sorry but I don't understand
Step-by-step explanation:
please forgive me
comment if I am forgiven
subtract 2x from both sides... 6x=2 divide by 6 = .3
Answer:
Step-by-step explanation:
The function y = x² + 16 is positive for any value of x, hence the graph has no intersection with the x-axis and therefore has no real zero's.
If no zero's, -4 also can't be its zero.