Question

If the inverse trigonometric identity ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \dfrac{{2\pi }}{3}$ holds true, then what will the value be of \[{\cot ^{ - 1}}x + {\cot ^{ - 1}}y\] is equal to
$
  (a){\text{ }}\dfrac{\pi }{2} \\
  (b){\text{ }}\dfrac{1}{2} \\
  (c){\text{ }}\dfrac{\pi }{3} \\
  (d){\text{ }}\dfrac{{\sqrt 3 }}{2} \\
  (e){\text{ }}\pi \\
$

Answer 1

Hint – In this question use the trigonometric identity that ${\tan ^{ - 1}}A + {\cot ^{ - 1}}A = \dfrac{\pi }{2}$, to change ${\tan ^{ - 1}}x{\text{ and ta}}{{\text{n}}^{ - 1}}y$ in the given equation into ${\cot ^{ - 1}}x{\text{ and co}}{{\text{t}}^{ - 1}}y$. This will help getting the answer.

Complete step-by-step solution -
Given trigonometric equation is
${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \dfrac{{2\pi }}{3}$
Now as we know that
${\tan ^{ - 1}}A + {\cot ^{ - 1}}A = \dfrac{\pi }{2}$
$ \Rightarrow {\tan ^{ - 1}}A = \dfrac{\pi }{2} - {\cot ^{ - 1}}A$ so use this property in above equation we have,
$ \Rightarrow \dfrac{\pi }{2} - {\cot ^{ - 1}}x + \dfrac{\pi }{2} - {\cot ^{ - 1}}y = \dfrac{{2\pi }}{3}$
Now simplify the above equation we have,
\[ \Rightarrow \dfrac{\pi }{2} + \dfrac{\pi }{2} - \dfrac{{2\pi }}{3} = {\cot ^{ - 1}}x + {\cot ^{ - 1}}y\]
\[ \Rightarrow {\cot ^{ - 1}}x + {\cot ^{ - 1}}y = \pi - \dfrac{{2\pi }}{3}\]
\[ \Rightarrow {\cot ^{ - 1}}x + {\cot ^{ - 1}}y = \dfrac{{3\pi - 2\pi }}{3} = \dfrac{\pi }{3}\]
So this is the required answer.
Hence option (C) is correct.

Note – As we have some basic trigonometric identities like ${\sin ^2}x + {\cos ^2}x = 1{\text{ and 1 + ta}}{{\text{n}}^2}x = {\sec ^2}x$, in the similar way we have identities involving inverse trigonometric ratios like ${\tan ^{ - 1}}A + {\cot ^{ - 1}}A = \dfrac{\pi }{2}{\text{ and si}}{{\text{n}}^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$. It is advised to remember these basic identities as it helps save a lot of time.