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

Last updated date: 02nd Aug 2024
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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.

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 $x=\dfrac{\pi }{4}$ .