Question

Prove that
\[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\]

Answer 1

Hint: First expand the given expression in left hand side using the formula for expansion of \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\]now 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 \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\]
We know that the formula for \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\]is given by \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]
Now applying the above formula we will get,
\[={{\tan }^{-1}}\left( \dfrac{(x)+\left( \dfrac{2x}{1-{{x}^{2}}} \right)}{1-\left( x \right)\left( \dfrac{2x}{1-{{x}^{2}}} \right)} \right)\]. . . . . . . . . . . . . . . . . . . . . . . .(1)
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{x-{{x}^{3}}+2x}{1-{{x}^{2}}}}{\dfrac{1-{{x}^{2}}-2{{x}^{2}}}{1-{{x}^{2}}}} \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2)
\[={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Hence proved that left hand side is equal to right hand side
Hence proved that \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\]

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