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Question: Prove that \[2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=...

Prove that 2tan⁡−1(34)−tan⁡−1(1731)=π42{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \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}}y and 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(34)−tan⁡−1(1731)2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \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)
Substituting x=34x=\dfrac{3}{4}in (1) we get,
=tan⁡−1(2×341−(34)2)−tan⁡−1(1731)={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{3}{4}}{1-{{\left( \dfrac{3}{4} \right)}^{2}}} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)
=tan⁡−1(64716)−tan⁡−1(1731)={{\tan }^{-1}}\left( \dfrac{\dfrac{6}{4}}{\dfrac{7}{16}} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)
=tan⁡−1(247)−tan⁡−1(1731)={{\tan }^{-1}}\left( \dfrac{24}{7} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Now (a) is in the form of tan⁡−1x−tan⁡−1y{{\tan }^{-1}}x-{{\tan }^{-1}}y
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=247x=\dfrac{24}{7}and y=1731y=\dfrac{17}{31}in (2) we get,
=tan⁡−1((247)−(1731)1+(247)(1731))={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{24}{7} \right)-\left( \dfrac{17}{31} \right)}{1+\left( \dfrac{24}{7} \right)\left( \dfrac{17}{31} \right)} \right)
=tan⁡−1(744−119217217+408217)={{\tan }^{-1}}\left( \dfrac{\dfrac{744-119}{217}}{\dfrac{217+408}{217}} \right)
=tan⁡−1(625217625217)={{\tan }^{-1}}\left( \dfrac{\dfrac{625}{217}}{\dfrac{625}{217}} \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. Also remember the fact that 2tan⁡−1(x)=tan⁡−1(2x1−x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) formula is nothing but obtained from tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y where yy is nothing but xx, therefore we can use both.