Question

Prove that \[{{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{4}{3} \right)\]

Answer 1

Hint: First expand the given expression in the left hand side using the formula for expansion of \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\]. Now expand the obtained equation using the formula \[2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\]the we will get the required expression in the right hand side.

Complete step-by-step answer:

Now take left hand side that is \[{{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)\]
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 by applying the above formula we will get,
\[={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{1}{4} \right)+\left( \dfrac{2}{9} \right)}{1-\left( \dfrac{1}{4} \right)\left( \dfrac{2}{9} \right)} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{9+8}{36}}{\dfrac{36-2}{36}} \right)\]
\[={{\tan }^{-1}}\left( \dfrac{17}{34} \right)\]
=\[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Multiplying and dividing with two to the above expression we will get
\[=\dfrac{1}{2}\left( 2{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
\[=\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{2}}{1-\dfrac{1}{4}} \right) \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
\[=\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{1}{\dfrac{3}{4}} \right) \right)\]
\[=\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{4}{3} \right) \right)\]
Hence prove that left hand side = right hand side
Hence proved that \[{{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{4}{3} \right)\]

Note: If \[xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]and if \[xy>1,{{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]. Important step that we have to remember here is that we have to multiply and divide the obtained LHS expression to get it similar to RHS .