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Question: Prove that \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\...

Prove that
tan⁡−1x+tan⁡−1(2x1−x2)=tan⁡−1(3x−x31−3x2){{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)

Explanation

Solution

Hint: First expand the given expression in left hand side using the formula for expansion of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}ynow substitute the values of x , y according to given expression and do the basic mathematical operations like addition and multiplication to get the required expression in the right hand side

Complete step-by-step answer:

First take the left hand side that is tan⁡−1x+tan⁡−1(2x1−x2){{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)
We know that the formula for tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}yis given by tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
Now applying the above formula we will get,
=tan⁡−1((x)+(2x1−x2)1−(x)(2x1−x2))={{\tan }^{-1}}\left( \dfrac{(x)+\left( \dfrac{2x}{1-{{x}^{2}}} \right)}{1-\left( x \right)\left( \dfrac{2x}{1-{{x}^{2}}} \right)} \right). . . . . . . . . . . . . . . . . . . . . . . .(1)
=tan⁡−1(x−x3+2x1−x21−x2−2x21−x2)={{\tan }^{-1}}\left( \dfrac{\dfrac{x-{{x}^{3}}+2x}{1-{{x}^{2}}}}{\dfrac{1-{{x}^{2}}-2{{x}^{2}}}{1-{{x}^{2}}}} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2)
=tan⁡−1(3x−x31−3x2)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Hence proved that left hand side is equal to right hand side
Hence proved that tan⁡−1x+tan⁡−1(2x1−x2)=tan⁡−1(3x−x31−3x2){{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)

Note: If xy<1,tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)and if xy>1,tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)xy>1,{{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right).Since the trigonometric functions are periodic functions, these functions are not bijections in their natural domains. We have used this formula xy<1,tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) because if we solve xy we will get the value < 1.