The correct options are
A 4x−2y=π+2
C 4x+2y=π+2
D y = 0
Put x=π4 in the given equation,
y=y3⇒y(y2−1)=0⇒y=−1,0,1
So the points are (π4,−1),(π4,0) and (π4,1)
Now given equation may be written as, y=tan2x.y3
Differentiating w.r.t x
dydx=tan2x.3y2dydx+y3.2tanx.sec2x
⇒dydx=2y3sec2x.tanx1−3tan2x.y2
Thus slope of tangents are,
m1=(dydx)(π4,−1)=2(−1).2.11−3.1.1=2
m2=(dydx)(π4,0)=2(0).2.11−3.1.0=0
m3=(dydx)(π4,1)=2(1).2.11−3.1.1=−2
Hence equation of tangents are,
(y+1)=2(x−π4)⇒4x−2y=π+2
(y−0)=0(x−π4)⇒y=0
(y−1)=2(x−π4)⇒4x+2y=π+2