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Question: Prove that \[2{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{...

Prove that 2tan⁡−1(12)+tan⁡−1(17)=tan⁡−1(3117)2{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{\tan }^{-1}}\left( \dfrac{31}{17} \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}}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. Use the formula 2tan⁡−1(x)=tan⁡−1(2x1−x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)

Complete step-by-step answer:

Now considering the L.H.S
L.H.S =
2tan⁡−1(12)+tan⁡−1(17)2{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
We know that
2tan⁡−1(x)=tan⁡−1(2x1−x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now substituting the value of x=12x=\dfrac{1}{2}in (1) we get,
=tan⁡−1(2×121−(12)2)={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{2}}{1-{{\left( \dfrac{1}{2} \right)}^{2}}} \right) +tan⁡−1(17)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
=tan⁡−1(11−(14))={{\tan }^{-1}}\left( \dfrac{1}{1-\left( \dfrac{1}{4} \right)} \right) +tan⁡−1(17)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
=tan⁡−1(134)={{\tan }^{-1}}\left( \dfrac{1}{\dfrac{3}{4}} \right) +tan⁡−1(17)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
=tan⁡−1(43)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) +tan⁡−1(17)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right) . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . (a)
(a) is in the form of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y.
Now applying the formula,
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
By substituting the values in (2) we get,
tan⁡−1((43)+(17)1−(43)(17)){{\tan }^{-1}}\left( \dfrac{\left( \dfrac{4}{3} \right)+\left( \dfrac{1}{7} \right)}{1-\left( \dfrac{4}{3} \right)\left( \dfrac{1}{7} \right)} \right)
=tan⁡−1(28+32121−421)={{\tan }^{-1}}\left( \dfrac{\dfrac{28+3}{21}}{\dfrac{21-4}{21}} \right)
=tan⁡−1(3117)={{\tan }^{-1}}\left( \dfrac{31}{17} \right) R.H.S

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. Therefore the inverse function does not exist. By identifying the proper domains they are bijections and so an inverse function exists. Take care while doing calculations.