Question

1) Solve for $x$: ${{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}(3x)$
2) Prove that ${{\tan }^{-1}}\left( \dfrac{6x-8{{x}^{3}}}{1-12{{x}^{2}}} \right)-{{\tan }^{-1}}\left( \dfrac{4x}{1-4{{x}^{2}}} \right)={{\tan }^{-1}}2x;\left| 2x \right|<\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: In the given questions we have to find the values of $x$for which is satisfy the condition. For this first of all we have to transpose ${{\tan }^{-1}}x$to the right-hand side and solve. We can transpose any terms, but we transpose that terms for whish calculation is easy. As the difference between $(x+1)-(x-1)=2\text{ and }3x-x=2x$ hence we transpose ${{\tan }^{-1}}x$. For solving such question, we have to use the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$
It should be noted that the above both formula valid if $xy<1$, in the first formula and $xy>-1$ in the second formula.

Complete step by step answer:
From question we have
${{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}(3x)$
Now transposing ${{\tan }^{-1}}x$to the right-hand side we can write
${{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}(3x)-{{\tan }^{-1}}(x)$
Now here we have to use the formula
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$
We can write further
$\begin{align}
  & {{\tan }^{-1}}\left( x-1 \right)+{{\tan }^{-1}}\left( x+1 \right)={{\tan }^{-1}}\left( \dfrac{x-1+x+1}{1-\left( x-1 \right)\left( x+1 \right)} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( x-1 \right)+{{\tan }^{-1}}\left( x+1 \right)={{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right)-----(1) \\
\end{align}$
Now we can write right-hand side as
$\begin{align}
  & {{\tan }^{-1}}(3x)-{{\tan }^{-1}}(x)={{\tan }^{-1}}\left( \dfrac{3x-x}{1+(3x)(x)} \right) \\
 & {{\tan }^{-1}}(3x)-{{\tan }^{-1}}(x)={{\tan }^{-1}}\left( \dfrac{2x}{1+3{{x}^{2}}} \right)-----(2) \\
\end{align}$
Now from $(1)\text{ and }(2)$we can write further
${{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+3{{x}^{2}}} \right)$
Now we take tangent to the both side we can write
$\tan \left\{ {{\tan }^{-1}}\left( \dfrac{2x}{2-{{x}^{2}}} \right) \right\}=\tan \left\{ {{\tan }^{-1}}\left( \dfrac{2x}{1+3{{x}^{2}}} \right) \right\}$
As we know that $\tan ({{\tan }^{-1}}x)=x$, so we can write further
$\dfrac{2x}{2-{{x}^{2}}}=\dfrac{2x}{1+3{{x}^{2}}}$
Further we can write
$\begin{align}
  & 2-{{x}^{2}}=1+3{{x}^{2}} \\
 & \Rightarrow 4{{x}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}=\dfrac{1}{2} \\
 & \Rightarrow x=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}$
Here we take $x=\dfrac{+1}{\sqrt{2}}$and ignore the negative value as ${{\tan }^{-1}}(x-1)$is negative.
Now for the second part we have
${{\tan }^{-1}}\left( \dfrac{6x-8{{x}^{3}}}{1-12{{x}^{2}}} \right)-{{\tan }^{-1}}\left( \dfrac{4x}{1-4{{x}^{2}}} \right)={{\tan }^{-1}}2x;\left| 2x \right|<\dfrac{1}{\sqrt{3}}$
Now we take left-hand side
${{\tan }^{-1}}\left( \dfrac{6x-8{{x}^{3}}}{1-12{{x}^{2}}} \right)-{{\tan }^{-1}}\left( \dfrac{4x}{1-4{{x}^{2}}} \right)$
Here we substitute $2x=\tan \theta $
So, we can write
${{\tan }^{-1}}\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right)-{{\tan }^{-1}}\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)$
As we know that
$\tan 3\theta =\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right)\text{ and tan2}\theta \text{=}\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)$
So, using the above formula we can write
\[\begin{align}
  & {{\tan }^{-1}}\left( \tan 3\theta \right)-{{\tan }^{-1}}\left( \tan 2\theta \right) \\
 & =3\theta -2\theta \\
 & =\theta \\
\end{align}\]
Now we put the value of $\theta $we can write
$\theta ={{\tan }^{-1}}2x$
Hence, we can write
${{\tan }^{-1}}\left( \dfrac{6x-8{{x}^{3}}}{1-12{{x}^{2}}} \right)-{{\tan }^{-1}}\left( \dfrac{4x}{1-4{{x}^{2}}} \right)={{\tan }^{-1}}2x$
Hence, second part of the question is proved.

Note:
When we have to solve the inverse trigonometric function, we must check that the given function is invertible or not in the given Domain.
A one-one onto function is called an invertible function.
If $f:X\to Y$be a one-one onto function, then each $y\in Y$, there exist unique element $x\in X$such that $f(x)=y$