If we take the square of x and square of y and then subtract them:
(csc t)²-(cot t)²=1 ( this eq. gets from basic identity
x²-y²=1......a 1+cot²x=csc²x)
equation 'a' represent the equation of hyperbola which is (x²/a²)-(y²/b²) =1 with given conditions( a=1,b=1)
So, option D is correct
X = -2
3 x -2 = -6
-6 -2= -8
|-8| = 8
-2 + 1= -1
|-1|=1
Not sure if I'm right but I believe they intersect at (1,4).
