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Question: If ab > -1, bc > -1 and ca > -1, then the value of \({\cot ^{ - 1}}\left( {\dfrac{{ab + 1}}{{a - ...

If ab > -1, bc > -1 and ca > -1, then the value of
cot⁡−1(ab+1a−b)+cot⁡−1(bc+1b−c)+cot⁡−1(ca+1c−a){\cot ^{ - 1}}\left( {\dfrac{{ab + 1}}{{a - b}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{bc + 1}}{{b - c}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{ca + 1}}{{c - a}}} \right) is
(A)−1\left( A \right) - 1
(B)cot⁡−1(a+b+c)\left( B \right){\cot ^{ - 1}}\left( {a + b + c} \right)
(C)cot⁡−1(abc)\left( C \right){\cot ^{ - 1}}\left( {abc} \right)
(D)0\left( D \right)0
(E)tan⁡−1(a+b+c)\left( E \right){\tan ^{ - 1}}\left( {a + b + c} \right)

Explanation

Solution

Hint– In this particular question use the concept that (cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}}), (tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)) and (tan⁡−1(−θ)=−tan⁡−1(θ){\tan ^{ - 1}}\left( { - \theta } \right) = - {\tan ^{ - 1}}\left( \theta \right)) so use these basic properties of trigonometry to reach the solution of the given problem.

Complete step-by-step answer:
Given equation is
cot⁡−1(ab+1a−b)+cot⁡−1(bc+1b−c)+cot⁡−1(ca+1c−a){\cot ^{ - 1}}\left( {\dfrac{{ab + 1}}{{a - b}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{bc + 1}}{{b - c}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{ca + 1}}{{c - a}}} \right)
Consider x=cot⁡−1Ax = {\cot ^{ - 1}}A
Now shift cot inverse to L.H.S
⇒cot⁡x=A\Rightarrow \cot x = A
Now as we know that cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}} so, substitute this value in above equation we have,
⇒1tan⁡x=A\Rightarrow \dfrac{1}{{\tan x}} = A
⇒tan⁡x=1A\Rightarrow \tan x = \dfrac{1}{A}
Now shift tan to R.H.S
⇒x=tan⁡−11A\Rightarrow x = {\tan ^{ - 1}}\dfrac{1}{A}
⇒cot⁡−1A=tan⁡−11A\Rightarrow {\cot ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{1}{A} So, use this property in given equation we have,
⇒tan⁡−1(a−bab+1)+tan⁡−1(b−cbc+1)+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{a - b}}{{ab + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{b - c}}{{bc + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now as we know that tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right) so, use this property in above equation we have,
⇒tan⁡−1(a−bab+1+b−cbc+11−(a−bab+1×b−cbc+1))+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{a - b}}{{ab + 1}} + \dfrac{{b - c}}{{bc + 1}}}}{{1 - \left( {\dfrac{{a - b}}{{ab + 1}} \times \dfrac{{b - c}}{{bc + 1}}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now simplify the above equation we have,
⇒tan⁡−1((a−b)(bc+1)+(b−c)(ab+1)(ab+1)(bc+1)−(a−b)(b−c))+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\left( {a - b} \right)\left( {bc + 1} \right) + \left( {b - c} \right)\left( {ab + 1} \right)}}{{\left( {ab + 1} \right)\left( {bc + 1} \right) - \left( {a - b} \right)\left( {b - c} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now again simplify the above equation we have,
⇒tan⁡−1(abc+a−b2c−b+ab2+b−abc−cab2c+ab+bc+1−ab+ac+b2−bc)+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{abc + a - {b^2}c - b + a{b^2} + b - abc - c}}{{a{b^2}c + ab + bc + 1 - ab + ac + {b^2} - bc}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now cancel out the terms
⇒tan⁡−1(a−b2c+ab2−cab2c+1+ac+b2)+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{a - {b^2}c + a{b^2} - c}}{{a{b^2}c + 1 + ac + {b^2}}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
⇒tan⁡−1((a−c)+b2(a−c)(1+b2)+ac(1+b2))+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\left( {a - c} \right) + {b^2}\left( {a - c} \right)}}{{\left( {1 + {b^2}} \right) + ac\left( {1 + {b^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
⇒tan⁡−1((a−c)(1+b2)(1+b2)(1+ac))+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\left( {a - c} \right)\left( {1 + {b^2}} \right)}}{{\left( {1 + {b^2}} \right)\left( {1 + ac} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now cancel out the common terms we have
⇒tan⁡−1(a−cca+1)+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{a - c}}{{ca + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
⇒tan⁡−1(−(c−a)ca+1)+tan⁡−1(c−aca+1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - \left( {c - a} \right)}}{{ca + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now as we know tan⁡−1(−θ)=−tan⁡−1(θ){\tan ^{ - 1}}\left( { - \theta } \right) = - {\tan ^{ - 1}}\left( \theta \right) so, use this property in above equation we have,
⇒−tan⁡−1(c−aca+1)+tan⁡−1(c−aca+1)\Rightarrow - {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{c - a}}{{ca + 1}}} \right)
Now as we see both terms are same but opposite sign
⇒cot⁡−1(ab+1a−b)+cot⁡−1(bc+1b−c)+cot⁡−1(ca+1c−a)=0\Rightarrow {\cot ^{ - 1}}\left( {\dfrac{{ab + 1}}{{a - b}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{bc + 1}}{{b - c}}} \right) + {\cot ^{ - 1}}\left( {\dfrac{{ca + 1}}{{c - a}}} \right) = 0
Hence option (D) is correct.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric properties whatever is used in this question is all stated up, then first convert the given equation in the terms of tan then apply the formula of (tan⁡−1a+tan⁡−1b{\tan ^{ - 1}}a + {\tan ^{ - 1}}b) and simplify as above then further simplify using similar properties as above we will get the required answer.