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

Solve 3tan⁡−1x=tan⁡−1(3x−x31−3x2)3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)

Explanation

Solution

From the given question we have to solve the 3tan⁡−1x=tan⁡−1(3x−x31−3x2)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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right). First, we have to divide 3tan⁡−1x3{{\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
⇒3tan⁡−1x=tan⁡−1(3x−x31−3x2)\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 3tan⁡−1x3{{\tan }^{-1}}x as
⇒3tan⁡−1x=tan⁡−1x+tan⁡−1x+tan⁡−1x\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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right).
By adding we will get
⇒3tan⁡−1x=tan⁡−1x+tan⁡−1(x+x1−x2)\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
⇒3tan⁡−1x=tan⁡−1x+tan⁡−1(2x1−x2)\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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right).
By adding we will get
⇒3tan⁡−1x=tan⁡−1(2x+x−x31−x2(1−x2−2x21−x2))\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
⇒3tan⁡−1x=tan⁡−1(2x+x−x31−x2(1−x2−2x21−x2))\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
⇒3tan⁡−1x=tan⁡−1(3x−x31−3x2)\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
⇒tan⁡−1x+cot⁡−1x=π2 ⇒sin⁡−1x+cos⁡−1x=π2 ⇒sec⁡−1x+cosec−1x=π2 ⇒tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy) \begin{aligned} & \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{aligned}
Students should be very careful while doing the calculations.