Question

The value of \[{\tan ^{ - 1}}\dfrac{{\left[ {a - b} \right]}}{{\left[ {1 + ab} \right]}} + {\tan ^{ - 1}}\dfrac{{\left[ {b - c} \right]}}{{\left[ {1 + bc} \right]}} = \]
A.\[{\tan ^{ - 1}}a - {\tan ^{ - 1}}b\]
B.\[{\tan ^{ - 1}}a - {\tan ^{ - 1}}c\]
C.\[{\tan ^{ - 1}}b - {\tan ^{ - 1}}c\]
D.\[{\tan ^{ - 1}}c - {\tan ^{ - 1}}a\]

Answer 1

Hint: In the given question we will use the formula of \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] for which \[xy < 1\] . Similarly , we will use this formula \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] , for which \[xy > 1\] . So , here we have given variables in the angle , so we can consider both the cases .

Complete step-by-step answer:
Given : \[{\tan ^{ - 1}}\dfrac{{\left[ {a - b} \right]}}{{\left[ {1 + ab} \right]}} + {\tan ^{ - 1}}\dfrac{{\left[ {b - c} \right]}}{{\left[ {1 + bc} \right]}}\] ,
 CASE 1 : When \[xy < 1\] , we have
Now using the identity \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] , for both terms of \[\tan \] we get ,
\[ = {\tan ^{ - 1}}a - {\tan ^{ - 1}}b + {\tan ^{ - 1}}b - {\tan ^{ - 1}}c\] ,
Cancelling out the terms of \[{\tan ^{ - 1}}b\] we get ,
\[ = {\tan ^{ - 1}}a - {\tan ^{ - 1}}c\] .
Therefore , option (2) is the correct answer for the given question .

CASE 2 : When \[xy > 1\] , we have .
Now using the identity \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] , for both terms of \[\tan \] in the question we get ,
\[ = \pi + {\tan ^{ - 1}}a - {\tan ^{ - 1}}b + \pi + {\tan ^{ - 1}}b - {\tan ^{ - 1}}c\]
Cancelling out the terms of \[{\tan ^{ - 1}}b\] we get ,
\[ = 2\pi + {\tan ^{ - 1}}a - {\tan ^{ - 1}}c\] .
Here , case 2 is not the solution according to the question , as the expression \[ = 2\pi + {\tan ^{ - 1}}a - {\tan ^{ - 1}}c\] is not given in the option . So , case 1 is a required solution for the question . But it is also a solution to this question which makes it a complete solution .
So, the correct answer is “Option B”.

Note: The inverse trigonometric functions are also called the anti – trigonometric functions or even known as arcus functions or cyclometric functions . The inverse trigonometric functions of sine , cosine , tangent , cosecant , secant and cotangent are used to find the angle of a triangle from any of the trigonometric functions . Inverse \[\tan \] is the inverse function of the trigonometric ratio ‘tangent’ . It is used to calculate the angle by applying the tangent ratio of the angle .