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Question: Find the value of \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b\], where is \[a > 0,b > 0,ab > 1\] equal to : ...

Find the value of tan⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}b, where is a>0,b>0,ab>1a > 0,b > 0,ab > 1 equal to :
A) tan⁡−1(a+b1−ab){\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)
B) tan⁡−1(a+b1−ab)−π{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) - \pi
C) π+tan⁡−1(a+b1−ab)\pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)
D) none of these

Explanation

Solution

To solve first of all let us see what we have been given to find tan⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}b. So as we can see we are provided with the conditionsa>0,b>0,ab>1a > 0,b > 0,ab > 1. Here we are supposed to apply the formula of the tan⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}b and later we have to check how the given conditions of a>0,b>0,ab=1a > 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⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}bwhile keeping in note of the given conditions that are stated as a>0,b>0,ab>1a > 0,b > 0,ab > 1
As we know that

tan⁡−1u+tan⁡−1v=tan⁡−1(u+v1−uv){\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⁡−1u+tan⁡−1v{\tan ^{ - 1}}u + {\tan ^{ - 1}}v
Similarly, tan⁡−1u−tan⁡−1v=tan⁡−1(u−v1+uv){\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⁡−1u−tan⁡−1v{\tan ^{ - 1}}u - {\tan ^{ - 1}}v
Now using the identity of tan⁡−1u+tan⁡−1v=tan⁡−1(u+v1−uv){\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⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}bwe would get –

{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = $$$${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)
Now checking about the given conditions in the question and modifying the above resultant in the desired way that is as follows –
The given conditions are –
a>0,b>0,ab>1a > 0,b > 0,ab > 1
Here a is given a is greater than 00and also b is greater than 00is given .Now take a note that a second condition is also been given to us in the question that is the answer defining one and is
ab>1ab > 1which means ab is greater than 11 .By that means denominator in tan⁡−1(a+b1−ab){\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)becomes negative so as we know that {\tan ^{ - 1}}\left( { - x} \right)$$$$ = \pi + {\tan ^{ - 1}}\left( x \right)-(as it is a standard identity of the function of tan⁡−1(−x){\tan ^{ - 1}}\left( { - x} \right)in Inverse trigonometry ).
Now applying this {\tan ^{ - 1}}\left( { - x} \right)$$$$ = \pi + {\tan ^{ - 1}}\left( x \right)on tan⁡−1(a+b1−ab){\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) as it has also become negative due to denominator in the brackets of the functiontan⁡−1(a+b1−ab){\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) becoming negative So ultimately {\tan ^{ - 1}}a + {\tan ^{ - 1}}b = $$$$ = \pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)
Now the resultant final answer is {\tan ^{ - 1}}a + {\tan ^{ - 1}}b = $$$$ = \pi + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) which is equal to the option C that is π+tan⁡−1(a+b1−ab)\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.