Answer:
because your grandfather has the experience to do work
Copy and paste ur answer onto internet it will tell u
Hope this helps you out . Of solving for k then just set it equal to zero
Given:


To find:
The quadrant of the terminal side of
and find the value of
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,


Here cos is positive and sine is negative. So,
must be lies in Quadrant IV.
We know that,



It is only negative because
lies in Quadrant IV. So,

After substituting
, we get





Therefore, the correct option is B.