Q.
Statement I The simplest form of tan−1a2−x2x,∣x∣<a is sin−1ax. Statement II The simplest form of tan−1(a3−3ax23a2x−x3),a>0,3−a<x<3a is 3tan−1x
I. Let x=asinθ, then sin−1(ax)=θ tan−1a2−x2x=tan−1(a2−a2sin2θasinθ) =tan−1(a1−sin2θasinθ)=tan−1(cosθsinθ) (∵sin2x+cos2x=1⇒cosx=1−sin2x) =tan−1(tanθ)=θ=sin−1(ax)
II. Let x=atanθ ⇒ax=tanθ⇒θ=tan−1(ax).....(i) ∴tan−1(a3−3ax23a2x−x3)=tan−1[a3−3a(atanθ)23a2(atanθ)−(atanθ)3] =tan−1(a3(1−3tan2θ)a3(3tanθ−tan3θ)) =tan−1(1−3tan2θ3tanθ−tan3θ)=tan−1(tan3θ) (∵tan3θ=1−3tan2θ3tanθ−tan3θ) =3θ=3tan−1(ax)[from Eq.(i)]