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

Prove that 2tan⁡−134−tan⁡−11731=π42{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}=\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 formulas tan⁡−1A−tan⁡−1B=tan⁡−1A−B1+AB{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A-B}{1+AB} and 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⁡−134−tan⁡−117312{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}
=tan⁡−134+tan⁡−134−tan⁡−11731={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}
Now, we know 34×34<1\dfrac{3}{4}\times \dfrac{3}{4}<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⁡−134+341−3×34×4−tan⁡−11731{{\tan }^{-1}}\dfrac{\dfrac{3}{4}+\dfrac{3}{4}}{1-\dfrac{3\times 3}{4\times 4}}-{{\tan }^{-1}}\dfrac{17}{31}
=tan⁡−13216−916−tan⁡−11731={{\tan }^{-1}}\dfrac{\dfrac{3}{2}}{\dfrac{16-9}{16}}-{{\tan }^{-1}}\dfrac{17}{31}
=tan⁡−1247−tan⁡−11731={{\tan }^{-1}}\dfrac{24}{7}-{{\tan }^{-1}}\dfrac{17}{31}
Now we know that tan⁡−1A−tan⁡−1B=tan⁡−1A−B1+AB{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A-B}{1+AB} .
tan⁡−1247−tan⁡−11731=tan⁡−1(247−17311+24×177×31){{\tan }^{-1}}\dfrac{24}{7}-{{\tan }^{-1}}\dfrac{17}{31}={{\tan }^{-1}}\left( \dfrac{\dfrac{24}{7}-\dfrac{17}{31}}{1+\dfrac{24\times 17}{7\times 31}} \right)
Now further solving the right-hand side of the above equation, we get
=tan⁡−1(24×31−7×177×317×13−24×177×31)={{\tan }^{-1}}\left( \dfrac{\dfrac{24\times 31-7\times 17}{7\times 31}}{\dfrac{7\times 13-24\times 17}{7\times 31}} \right)
=tan⁡−1(744−119217+408)={{\tan }^{-1}}\left( \dfrac{744-119}{217+408} \right)
=tan⁡−1(625625)={{\tan }^{-1}}\left( \dfrac{625}{625} \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⁡−134−tan⁡−11731=π42{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}=\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] . Also it is important to check whether the multiplication of A and B is less than one or not to use proper formula.