Question

Prove that $2{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}=\dfrac{\pi }{4}$ .

Answer 1

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

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.
$2{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}$
$=ta{{n}^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}$
Now, we know $\dfrac{1}{3}\times \dfrac{1}{3}<1$ . So, if we use the formula ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB}$ , we get
${{\tan }^{-1}}\dfrac{\dfrac{1}{3}+\dfrac{1}{3}}{1-\dfrac{1}{3\times 3}}+{{\tan }^{-1}}\dfrac{1}{7}$
$={{\tan }^{-1}}\dfrac{\dfrac{2}{3}}{\dfrac{9-1}{9}}+{{\tan }^{-1}}\dfrac{1}{7}$
$={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{1}{7}$
Now, we know $\dfrac{3}{4}\times \dfrac{1}{7}<1$ . So, if we use the formula ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB}$ , we get
${{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3\times 1}{4\times 7}} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{4\times 7-3\times 1}{4\times 7}} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} \right)$
$={{\tan }^{-1}}1$
We know that the value of ${{\tan }^{-1}}1$ is equal to $\dfrac{\pi }{4}$ , 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 $2{{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{7}=\dfrac{\pi }{4}$ .

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 ${{\sin }^{-1}}x$ is $[-1,1]$ and the range is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ .