Question

Prove the following result of inverse trigonometric functions:
 $ {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)=\left\{ \begin{matrix}
   0,\text{ if }\dfrac{\pi }{4} < x < \dfrac{\pi }{2} \\
   \pi ,\text{ if }0 < x < \dfrac{\pi }{4} \\
\end{matrix} \right. $ .

Answer 1

Hint:
We start solving the problem by recalling the facts that $ 0 < \tan x < 1 $ , for $ 0 < x < \dfrac{\pi }{4} $ and $ 1 < \tan x < \infty $ , for $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ , $ 0 < \cot x < 1 $ , for $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ and \[1 < \cot x < \infty \], for $ 0 < x < \dfrac{\pi }{4} $ . We then consider the interval $ 0 < x < \dfrac{\pi }{4} $ and make use of the results $ {{\tan }^{-1}}a+{{\tan }^{-1}}b=\left\{ \begin{matrix}
   {{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab < 1 \\
   \pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b > 0 \\
   -\pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b < 0 \\
\end{matrix} \right. $ to proceed through the problem. We then make the necessary calculations and make use of the results $ \cot x=\dfrac{1}{\tan x} $ , $ \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} $ and $ {{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right) $ to get the required answer. We follow similar procedure by considering the interval $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ to complete the given result.

Complete step by step answer:
According to the problem, we are asked to prove the given result: $ {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)=\left\{ \begin{matrix}
   0,\text{ if }\dfrac{\pi }{4} < x < \dfrac{\pi }{2} \\
   \pi ,\text{ if }0 < x < \dfrac{\pi }{4} \\
\end{matrix} \right. $ .
We know that $ 0 < \tan x < 1 $ , for $ 0 < x < \dfrac{\pi }{4} $ and $ 1 < \tan x < \infty $ , for $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ .
Also $ 0 < \cot x < 1 $ , for $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ and \[1 < \cot x < \infty \], for $ 0 < x < \dfrac{\pi }{4} $ .
Let us take the interval $ 0 < x < \dfrac{\pi }{4} $ .
Now, let us consider $ {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right) $ .
We know that $ {{\tan }^{-1}}a+{{\tan }^{-1}}b=\left\{ \begin{matrix}
   {{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab < 1 \\
   \pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b > 0 \\
   -\pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b < 0 \\
\end{matrix} \right. $ .
So, we have \[1 < \cot x < \infty \], for $ 0 < x < \dfrac{\pi }{4} $ so, we get \[1 < {{\cot }^{3}}x < \infty \], for $ 0 < x < \dfrac{\pi }{4} $ . So, we get $ \left( \cot x \right)\left( {{\cot }^{3}}x \right) > 1 $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\cot A+{{\cot }^{3}}A}{1-\left( \cot A \right)\left( {{\cot }^{3}}A \right)} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\cot A+{{\cot }^{3}}A}{1-{{\cot }^{4}}A} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\cot x\left( 1+{{\cot }^{2}}A \right)}{\left( 1+{{\cot }^{2}}A \right)\left( 1-{{\cot }^{2}}A \right)} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\cot A}{1-{{\cot }^{2}}A} \right) $ .
We know that $ \cot x=\dfrac{1}{\tan x} $ and $ \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\times \left( \dfrac{2\tan A}{1-{{\tan }^{2}}A} \right) \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{\tan A}}{1-\dfrac{1}{{{\tan }^{2}}A}} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)+\pi +{{\tan }^{-1}}\left( \dfrac{\tan A}{{{\tan }^{2}}A-1} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)+\pi +{{\tan }^{-1}}\left( \dfrac{-\tan A}{1-{{\tan }^{2}}A} \right) $ .
We know that $ {{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)+\pi -{{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)=\pi $ , for $ 0 < x < \dfrac{\pi }{4} $ ---(1).
Now, let us consider the interval $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ .
So, we have $ 0 < \cot x < 1 $ , for $ 0 < x < \dfrac{\pi }{4} $ so, we get \[0 < {{\cot }^{3}}x < 1\], for $ 0 < x < \dfrac{\pi }{4} $ . So, we get $ \left( \cot x \right)\left( {{\cot }^{3}}x \right) < 1 $ .
We know that $ {{\tan }^{-1}}a+{{\tan }^{-1}}b=\left\{ \begin{matrix}
   {{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab < 1 \\
   \pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b > 0 \\
   -\pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right);ab > 1,a,b < 0 \\
\end{matrix} \right. $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \dfrac{\cot A+{{\cot }^{3}}A}{1-\left( \cot A \right)\left( {{\cot }^{3}}A \right)} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \dfrac{\cot A+{{\cot }^{3}}A}{1-{{\cot }^{4}}A} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \dfrac{\cot x\left( 1+{{\cot }^{2}}A \right)}{\left( 1+{{\cot }^{2}}A \right)\left( 1-{{\cot }^{2}}A \right)} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \dfrac{\cot A}{1-{{\cot }^{2}}A} \right) $ .
We know that $ \cot x=\dfrac{1}{\tan x} $ and $ \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{1}{2}\times \left( \dfrac{2\tan A}{1-{{\tan }^{2}}A} \right) \right)+{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{\tan A}}{1-\dfrac{1}{{{\tan }^{2}}A}} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)+{{\tan }^{-1}}\left( \dfrac{\tan A}{{{\tan }^{2}}A-1} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)+{{\tan }^{-1}}\left( \dfrac{-\tan A}{1-{{\tan }^{2}}A} \right) $ .
We know that $ {{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)={{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right)-{{\tan }^{-1}}\left( \dfrac{\tan A}{1-{{\tan }^{2}}A} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)=0 $ , for $ \dfrac{\pi }{4} < x < \dfrac{\pi }{2} $ ---(2).
From equations (1) and (2), we can see that we have proved the result $ {{\tan }^{-1}}\left( \dfrac{1}{2}\tan 2A \right)+{{\tan }^{-1}}\left( \cot A \right)+{{\tan }^{-1}}\left( {{\cot }^{3}}A \right)=\left\{ \begin{matrix}
   0,\text{ if }\dfrac{\pi }{4} < x < \dfrac{\pi }{2} \\
   \pi ,\text{ if }0 < x < \dfrac{\pi }{4} \\
\end{matrix} \right. $ .

Note:
We can see that the given problem contains a huge amount of calculation so, we need to perform each step carefully to avoid confusion. We should not always consider $ {{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right) $ , as this is not valid for every value of a and b. Whenever we get the problems involving a sum of inverse tangent terms, we should check the product of the terms present inside the inverse. Similarly, we can expect problem to check the continuity of function at $ x=0 $ and $ x=\dfrac{\pi }{4} $ .