Question

Solve $3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$

Answer 1

Hint: From the given question we have to solve the $3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$. To solve the above question, we should use the formulas of inverse trigonometry ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$. First, we have to divide $3{{\tan }^{-1}}x$ and then we have to apply the above formula and we will get the final result.

Complete step by step solution:
From the given question we have to solve the
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$
Now we have to rewrite the right-hand side part in the following manner $3{{\tan }^{-1}}x$ as
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}x+{{\tan }^{-1}}x+{{\tan }^{-1}}x$
Now we have to add the first two terms in the left-hand side part of the equation,
 As we know that the formula of inverse trigonometry ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$.
By adding we will get
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{x+x}{1-{{x}^{2}}} \right)$
By further simplifying the left hand side part of the equation we will get
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$
now again further we will add the terms in the left-hand side part of the equation,
As we know that the formula of inverse trigonometry ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)$.
By adding we will get
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{\dfrac{2x+x-{{x}^{3}}}{1-{{x}^{2}}}}{\left( \dfrac{1-{{x}^{2}}-2{{x}^{2}}}{1-{{x}^{2}}} \right)} \right)$

 By further simplifying the left-hand side part of the equation we will get
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{\dfrac{2x+x-{{x}^{3}}}{1-{{x}^{2}}}}{\left( \dfrac{1-{{x}^{2}}-2{{x}^{2}}}{1-{{x}^{2}}} \right)} \right)$
By further simplifying the left-hand side part of the equation we will get
$\Rightarrow 3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$
Therefore, left hand side is equal to the right-hand side hence proved.

Note: Students should know the basic formulas of inverse trigonometry. Students should know the formulas like
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\sec }^{-1}}x+\text{cose}{{c}^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right) \\
\end{align}$
Students should be very careful while doing the calculations.