Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
p (2) = 4×2^4-13×2^2-2×2
= 64 - 52 - 4
= 64 - 56
= 8
Therefore, remainder = 8
Given:

x lies in the III quadrant.
To find:
The values of
.
Solution:
It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.
We know that,




x lies in the III quadrant. So,


Now,



And,





We know that,



Therefore, the required values are
.
D is right because there is not tire