Question

What will be the value of ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b$, where a >0, b>0, ab>1.
$
  (a){\text{ ta}}{{\text{n}}^{ - 1}}\dfrac{{a + b}}{{1 - ab}} \\
  (b){\text{ ta}}{{\text{n}}^{ - 1}}\dfrac{{a + b}}{{1 - ab}} - \pi \\
  (c){\text{ }}\pi {\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{{a + b}}{{1 - ab}} \\
  (d){\text{ none of these}} \\
$

Answer 1

Hint: In this question use ab>1, to find that $\dfrac{{a + b}}{{1 - ab}}$ is greater than 0 or less than 0, thus check for ${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$, that it lies in which quadrant, then accordingly check which option is true.

Complete step-by-step answer:
Given trigonometric equation

${\tan ^{ - 1}}a + {\tan ^{ - 1}}b$ , where a > 0, b > 0 and ab > 1

As we know that

${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$

But in this formula ab < 1 and it lies in the first quadrant.

But it is given that ab > 1

Therefore (1 – ab) < 0.

Therefore $\dfrac{{a + b}}{{1 - ab}}$ < 0

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ Is negative angle.

Now as we know tan is positive in the first and third quadrant.

Therefore ${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ lies either in the second or fourth quadrant.

So we have to add either $\pi $ or $2\pi $ in ${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}}

\right)$ so that it lies either in the second quadrant or in the fourth quadrant.

Therefore
${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$

(Lies in second quadrant)

Or

${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = 2\pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ (Lies in fourth quadrant)
So according to options, option (C) is correct.

So this is the required answer.

Note: In the quadrant system $0{\text{ to }}\dfrac{\pi }{2}$ refers to first quadrant, the second quadrant lies between $\dfrac{\pi }{2}{\text{to}}\pi $, the third quadrant from $\pi {\text{ to }}\dfrac{{3\pi }}{2}$ and then the fourth quadrant from $\dfrac{{3\pi }}{2}{\text{ to 2}}\pi $. The tricky part here is that we have added $\pi $ with ${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ and told that it lies in second quadrant, according to the angle of quadrant measurement $\pi + \theta $ lies in third quadrant, it is because we have showed that ${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ is a negative angle, and thus it will eventually get subtracted despite of the “+” sign.