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Question

Prove that:
tan1(12tan2A)+tan1(cotA)+tan1(cot3A)=0,ifπ4<A<π2π,if0<A<π/4

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Solution

tan1(12tan2A)+tan1(cotA)+tan1(cot3A)=0, if π4<A<π2π, if 0<A<π4
we know that
tan1x+tan1y=⎪ ⎪ ⎪⎪ ⎪ ⎪tan1(x+y1xy) if xy<1π+tan1(x+y1+xy) if xy>1
Now, for 0<A<π4cotA>1 and for π4<a<π2cotA<1
we can with
tan1(cotA)+tan1(cot3A)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪tan1(cotA+cot3A1cot4A) if π4<A<π2π+tan1(cotA+cot3A1cot4A) if 0<A<π2
=⎪ ⎪ ⎪⎪ ⎪ ⎪tan1(cotA1cot2A) if π0<A<π2π+tan1(cotA1cot2A) if 0<A<π2
[As cotA+cot3A1cot4A=cotA(1+cot2A)(1+cot2A)(1+cot2A)]
Now cotA1cot2A=1tanθ11tan2A=tanAtan2A1=12tan2A
tan1(cotA)+tan1(cot3A)=⎪ ⎪ ⎪⎪ ⎪ ⎪tan1(12tan2A) if π4<A<π2π+tan1(12tan2A) if 0<A<π2
=⎪ ⎪ ⎪⎪ ⎪ ⎪tan1(12tan2A) if π4<A<π2πtan1(12tan2A) if 0<A<π2
Adding tan1(12tan2A) or both sides we get
tan1(12tan2A)+tan1(cotA)+tan1(cot3A)=0 if π4<A<π2π if 0<A<π2

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