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Question: Prove that the expression \({\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}\...

Prove that the expression tan⁡−1x+cot⁡−1(x+1)=tan⁡−1(x2+x+1){\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}\left( {{x^2} + x + 1} \right).

Explanation

Solution

Hint: We need to prove that in the given expression the left hand side is equal to the right hand side. Use the basic inverse trigonometric identities involving tan⁡−1x{\tan ^{ - 1}}x and cot⁡−1(x+1){\cot ^{ - 1}}\left( {x + 1} \right) such that sum of these two is equal to π2\dfrac{\pi }{2} along with the basic formula involving addition and subtraction of two tan⁡−1entities{\tan ^{ - 1}}{\text{entities}} to get the proof.

Complete step-by-step answer:
Given equation is
tan⁡−1x+cot⁡−1(x+1)=tan⁡−1(x2+x+1){\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}\left( {{x^2} + x + 1} \right)
Now consider L.H.S
⇒tan⁡−1x+cot⁡−1(x+1)\Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right)……….. (1)
As we know that cot⁡−1A+tan⁡−1A=π2{\cot ^{ - 1}}A + {\tan ^{ - 1}}A = \dfrac{\pi }{2}.
⇒cot⁡−1A=π2−tan⁡−1A\Rightarrow {\cot ^{ - 1}}A = \dfrac{\pi }{2} - {\tan ^{ - 1}}A
So, use this property in equation (1) we have,
⇒tan⁡−1x+π2−tan⁡−1(x+1)\Rightarrow {\tan ^{ - 1}}x + \dfrac{\pi }{2} - {\tan ^{ - 1}}\left( {x + 1} \right)
=tan⁡−1x−tan⁡−1(x+1)+π2= {\tan ^{ - 1}}x - {\tan ^{ - 1}}\left( {x + 1} \right) + \dfrac{\pi }{2}
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(x−x−11+x(x+1))+π2\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{x - x - 1}}{{1 + x\left( {x + 1} \right)}}} \right) + \dfrac{\pi }{2}
⇒tan⁡−1(−11+x+x2)+π2\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{1 + x + {x^2}}}} \right) + \dfrac{\pi }{2}……….. (2)
Now as we know tan⁡−1A−tan⁡−1(−1A)=π2{\tan ^{ - 1}}A - {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = \dfrac{\pi }{2}
⇒tan⁡−1(−1A)=tan⁡−1A−π2\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = {\tan ^{ - 1}}A - \dfrac{\pi }{2} So, use this property in equation (2) we have,
⇒tan⁡−1(−11+x+x2)+π2=tan⁡−1(1+x+x2)−π2+π2\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{1 + x + {x^2}}}} \right) + \dfrac{\pi }{2} = {\tan ^{ - 1}}\left( {1 + x + {x^2}} \right) - \dfrac{\pi }{2} + \dfrac{\pi }{2}
=tan⁡−1(1+x+x2)= {\tan ^{ - 1}}\left( {1 + x + {x^2}} \right)
= R.H.S
Hence Proved.
Note: Whenever we face such proving questions involving trigonometric identities the key point is simply to have the understanding of basic inverse trigonometric identities, some of them are mentioned above. The knowledge of these identities will help you get on the right track to reach the answer.