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Question: Prove that \[2{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)=\d...

Prove that 2tan⁡−1(13)+tan⁡−1(17)=π42{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)=\dfrac{\pi }{4}

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:
Now considering L.H.S,
2tan⁡−1(13)+tan⁡−1(17)2{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
The first term is in the form of 2tan⁡−1(x)2{{\tan }^{-1}}\left( x \right)
Now applying the formula,
2tan⁡−1(x)=tan⁡−1(2x1−x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting x=13x=\dfrac{1}{3}in (1) we get
=tan⁡−1(2×131−(13)2)+tan⁡−1(17)={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{3}}{1-{{\left( \dfrac{1}{3} \right)}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
=tan⁡−1(2389)+tan⁡−1(17)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
=tan⁡−1(34)+tan⁡−1(17)={{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
The above expression (a) is in the form of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y
By 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)
Substituting x=34x=\dfrac{3}{4}and y=17y=\dfrac{1}{7}in (2) we get,
=tan⁡−1((34)+(17)1−(34)(17))={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{3}{4} \right)+\left( \dfrac{1}{7} \right)}{1-\left( \dfrac{3}{4} \right)\left( \dfrac{1}{7} \right)} \right)
=tan⁡−1(21+42828−328)={{\tan }^{-1}}\left( \dfrac{\dfrac{21+4}{28}}{\dfrac{28-3}{28}} \right)
=tan⁡−1(25282528)={{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} \right)
=tan⁡−1(1)={{\tan }^{-1}}\left( 1 \right)
=π4=\dfrac{\pi }{4}
= 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.