Question

Find the value of \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\], where is \[a > 0,b > 0,ab > 1\] equal to :
A) \[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
B) \[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) - \pi \]
C) \[\pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
D) none of these

Answer 1

Hint:
To solve first of all let us see what we have been given to find \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\]. So as we can see we are provided with the conditions\[a > 0,b > 0,ab > 1\]. Here we are supposed to apply the formula of the \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\] and later we have to check how the given conditions of \[a > 0,b > 0,ab = 1\] are satisfied to get the accurate and the desired answer.

Complete step by step solution:
Now we are here asked to find the \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\]while keeping in note of the given conditions that are stated as \[a > 0,b > 0,ab > 1\]
As we know that

\[{\tan ^{ - 1}}u + {\tan ^{ - 1}}v = {\tan ^{ - 1}}\left( {\dfrac{{u + v}}{{1 - uv}}} \right)\] which is an identity of inverse trigonometry function of \[{\tan ^{ - 1}}u + {\tan ^{ - 1}}v\]
Similarly, \[{\tan ^{ - 1}}u - {\tan ^{ - 1}}v = {\tan ^{ - 1}}\left( {\dfrac{{u - v}}{{1 + uv}}} \right)\] which is also an identity of inverse trigonometry function of \[{\tan ^{ - 1}}u - {\tan ^{ - 1}}v\]
Now using the identity of \[{\tan ^{ - 1}}u + {\tan ^{ - 1}}v = {\tan ^{ - 1}}\left( {\dfrac{{u + v}}{{1 - uv}}} \right)\] upon the terms asked in the question that is \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\]we would get –

\[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \]\[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
Now checking about the given conditions in the question and modifying the above resultant in the desired way that is as follows –
The given conditions are –
\[a > 0,b > 0,ab > 1\]
Here a is given a is greater than \[0\]and also b is greater than \[0\]is given .Now take a note that a second condition is also been given to us in the question that is the answer defining one and is
\[ab > 1\]which means ab is greater than \[1\] .By that means denominator in \[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]becomes negative so as we know that \[{\tan ^{ - 1}}\left( { - x} \right)\]\[ = \pi + {\tan ^{ - 1}}\left( x \right)\]-(as it is a standard identity of the function of \[{\tan ^{ - 1}}\left( { - x} \right)\]in Inverse trigonometry ).
Now applying this \[{\tan ^{ - 1}}\left( { - x} \right)\]\[ = \pi + {\tan ^{ - 1}}\left( x \right)\]on \[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] as it has also become negative due to denominator in the brackets of the function\[{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] becoming negative So ultimately \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \]\[ = \pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
Now the resultant final answer is \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \]\[ = \pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] which is equal to the option C that is \[\pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] among the given options

At last the option C is the correct answer.

Note:
While approaching such kinds of questions that are mainly based upon the application of the basic identities of the inverse trigonometry function. One should be equipped with the knowledge of these identities and the conditions and the different modifications in different those conditions to solve this type of questions easily and accurately.