If tan−1⁡y=4tan−1⁡x, then 1/y is zero for

If we put x = tanθ, the given equality becomes tan–1y = 4θ⇒y=tan⁡4θ=2tan⁡2θ1−tan2⁡2θ=22tan⁡θ1−tan2⁡θ1−2tan⁡θ1−tan2⁡θ2=2×2x1−x21−x22−4x2=4x1−x21−6x2+x4∴1y is zero if x4−6x2+1=0⇒x2=6±36−42=3±22=(1±2)2