tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A)=⎧⎪⎨⎪⎩0, if π4<A<π2π, if 0<A<π4
we know that
tan−1x+tan−1y=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩tan−1(x+y1−xy) if xy<1π+tan−1(x+y1+xy) if xy>1
Now, for 0<A<π4cotA>1 and for π4<a<π2cotA<1
∴ we can with
tan−1(cotA)+tan−1(cot3A)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩tan−1(cotA+cot3A1−cot4A) if π4<A<π2π+tan−1(cotA+cot3A1−cot4A) if 0<A<π2
=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩tan−1(cotA1−cot2A) if π0<A<π2π+tan−1(cotA1−cot2A) if 0<A<π2
[As cotA+cot3A1−cot4A=cotA(1+cot2A)(1+cot2A)(1+cot2A)]
Now cotA1−cot2A=1tanθ1−1tan2A=tanAtan2A−1=12tan2A
∴tan−1(cotA)+tan−1(cot3A)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩tan−1(−12tan2A) if π4<A<π2π+tan−1(−12tan2A) if 0<A<π2
=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩tan−1(12tan2A) if π4<A<π2π−tan−1(12tan2A) if 0<A<π2
Adding tan−1(12tan2A) or both sides we get
tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A)=⎧⎪⎨⎪⎩0 if π4<A<π2π if 0<A<π2