Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
Prove that
${tan}^{-1}\left(\frac{1}{2}tan2A\right)+{tan}^{-1}\left(cotA\right)+{tan}^{-1}\left({cot}^{3}A\right)$
$0$ if $\pi /4<A<\pi /2$
and $=\pi$ if $0<A<\pi /4$.

Answer 1

First note that $cotA>1$ if $0<A<\pi /4$

and $cotA<1$ if $\pi /4<A<\pi /2$ Hence

$\therefore$ ${tan}^{-1}\left(cotA\right)+{tan}^{-1}\left({cot}^{3}A\right)$

$=\pi +{tan}^{-1}\frac{cotA+{cot}^{3}A}{1-{cot}^{4}A}$ if $0<A<\frac{\pi }{4}$

and $={tan}^{-1}\frac{cotA+{cot}^{3}A}{1-{cot}^{4}A}$ if $\frac{\pi }{4}<A<\frac{\pi }{2}$.

Also $\frac{cotA{cot}^{3}A}{1-{cot}^{4}A}=\frac{cotA}{1-{cot}^{2}a}$

$=\frac{cosAsinA}{({sin}^{2}A-{cos}^{2}A)}$

$=-\frac{sin2A}{2cos2A}=-\frac{1}{2}tan2A$

Hence ${tan}^{-1}\left(\frac{1}{2}tan2A\right)+{tan}^{-1}\left(cotA\right)+{tan}^{-1}\left({cot}^{3}A\right)$

$=\pi$ in the first case,

$=0$ in the 2nd case because

${tan}^{-1}(-x)=-{tan}^{-1}x$