Prove that-

Question

tan^-1(1/2tan 2A)+tan^-1(cotA)+tan^-1(cot^3A)
={0,ifpi/4<A<pi/2
{pi,0<A<pi/4

Where 2 small curly brackets are 1 big curly bracket

Answer 1

Let
θ=tan^-1(cotA)
Ø=tan^-1(cot^3A)
Now, using the last two terms, we get
tan(θ+Ø) = (tanθ+tanØ)/(1-tanθtanØ)
= (cotA+cot^3A)/(1-cotAcot^3A)
= (cotA(1+cot^2A))/((1-cot^2A)(1+cot^2A))
= cotA/(1-cot^2A)
= -1/2 tan2A
That means that θ+Ø = arctan(-1/2 tan2A) and our original expression is just
tan^-1(1/2tan 2A) + tan^-1(-1/2 tan 2A)
= 0 or maybe π