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Question: The value of \[{\tan ^{ - 1}}\dfrac{{\left[ {a - b} \right]}}{{\left[ {1 + ab} \right]}} + {\tan ^{ ...

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

Explanation

Solution

Hint : In the given question we will use the formula of tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) for which xy<1xy < 1 . Similarly , we will use this formula tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) , for which xy>1xy > 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[a−b][1+ab]+tan⁡−1[b−c][1+bc]{\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<1xy < 1 , we have
Now using the identity tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) , for both terms of tan⁡\tan we get ,
=tan⁡−1a−tan⁡−1b+tan⁡−1b−tan⁡−1c= {\tan ^{ - 1}}a - {\tan ^{ - 1}}b + {\tan ^{ - 1}}b - {\tan ^{ - 1}}c ,
Cancelling out the terms of tan⁡−1b{\tan ^{ - 1}}b we get ,
=tan⁡−1a−tan⁡−1c= {\tan ^{ - 1}}a - {\tan ^{ - 1}}c .
Therefore , option (2) is the correct answer for the given question .

CASE 2 : When xy>1xy > 1 , we have .
Now using the identity tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) , for both terms of tan⁡\tan in the question we get ,
=π+tan⁡−1a−tan⁡−1b+π+tan⁡−1b−tan⁡−1c= \pi + {\tan ^{ - 1}}a - {\tan ^{ - 1}}b + \pi + {\tan ^{ - 1}}b - {\tan ^{ - 1}}c
Cancelling out the terms of tan⁡−1b{\tan ^{ - 1}}b we get ,
=2π+tan⁡−1a−tan⁡−1c= 2\pi + {\tan ^{ - 1}}a - {\tan ^{ - 1}}c .
Here , case 2 is not the solution according to the question , as the expression =2π+tan⁡−1a−tan⁡−1c = 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⁡\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 .