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Question: Prove that \(2{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}=\dfrac{\pi }{4}\) ....

Prove that 2tan⁡−113+tan⁡−117=π42{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}=\dfrac{\pi }{4} .

Explanation

Solution

Hint: The above question is related to inverse trigonometric function and for solving the problem, you need to use the formula tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} .

Complete step-by-step answer:
Now moving to the solution to the above question, we will start with the left-hand side of the equation given in the question.
2tan⁡−113+tan⁡−1172{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}
=tan−113+tan⁡−113+tan⁡−117=ta{{n}^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}
Now, we know 13×13<1\dfrac{1}{3}\times \dfrac{1}{3}<1 . So, if we use the formula tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} , we get
tan⁡−113+131−13×3+tan⁡−117{{\tan }^{-1}}\dfrac{\dfrac{1}{3}+\dfrac{1}{3}}{1-\dfrac{1}{3\times 3}}+{{\tan }^{-1}}\dfrac{1}{7}
=tan⁡−1239−19+tan⁡−117={{\tan }^{-1}}\dfrac{\dfrac{2}{3}}{\dfrac{9-1}{9}}+{{\tan }^{-1}}\dfrac{1}{7}
=tan⁡−134+tan⁡−117={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{1}{7}
Now, we know 34×17<1\dfrac{3}{4}\times \dfrac{1}{7}<1 . So, if we use the formula tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} , we get
tan⁡−1(34+171−3×14×7){{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3\times 1}{4\times 7}} \right)
=tan⁡−1(25284×7−3×14×7)={{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{4\times 7-3\times 1}{4\times 7}} \right)
=tan⁡−1(25282528)={{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} \right)
=tan⁡−11={{\tan }^{-1}}1
We know that the value of tan⁡−11{{\tan }^{-1}}1 is equal to π4\dfrac{\pi }{4} , which is equal to the right-hand side of the equation that we are asked to prove. So, we can say that we have proved that 2tan⁡−113+tan⁡−117=π42{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}=\dfrac{\pi }{4} .

Note: While dealing with inverse trigonometric functions, it is preferred to know about the domains and ranges of the different inverse trigonometric functions. For example: the domain of sin⁡−1x{{\sin }^{-1}}x is [−1,1][-1,1] and the range is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] .