If x<−13, then tan−13x−x31−3x2 equals
3tan−1x
−π+3tan−1x
π+3tan−1x
none of these
Let tan−1x=θ. Then, x=tanθ
Also,
x<−13⇒tanθ<−13⇒−π2<θ<−π6
Now,
tan−13x−x31−3x2= tan−1(tan3θ)= tan−1(tan(π+30))=π+3θ=π+3tan−1x