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Question: Prove that \[{{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\df...

Prove that tan⁡−1(14)+tan⁡−1(29)=12tan⁡−1(43){{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{4}{3} \right)

Explanation

Solution

Hint: First expand the given expression in the left hand side using the formula for expansion of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y. Now expand the obtained equation using the formula 2tan⁡−1x=tan⁡−1(2x1−x2)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(14)+tan⁡−1(29){{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \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 by applying the above formula we will get,
=tan⁡−1((14)+(29)1−(14)(29))={{\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(9+83636−236)={{\tan }^{-1}}\left( \dfrac{\dfrac{9+8}{36}}{\dfrac{36-2}{36}} \right)
=tan⁡−1(1734)={{\tan }^{-1}}\left( \dfrac{17}{34} \right)
=tan⁡−1(12){{\tan }^{-1}}\left( \dfrac{1}{2} \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Multiplying and dividing with two to the above expression we will get
=12(2tan⁡−1(12))=\dfrac{1}{2}\left( 2{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
=12(tan⁡−1(2×121−14))=\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{2}}{1-\dfrac{1}{4}} \right) \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
=12(tan⁡−1(134))=\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{1}{\dfrac{3}{4}} \right) \right)
=12(tan⁡−1(43))=\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(14)+tan⁡−1(29)=12tan⁡−1(43){{\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⁡−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). 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 .