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

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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 ${\tan ^{ - 1}}x$ and ${\cot ^{ - 1}}\left( {x + 1} \right)$ such that sum of these two is equal to $\dfrac{\pi }{2}$ along with the basic formula involving addition and subtraction of two ${\tan ^{ - 1}}{\text{entities}}$ to get the proof.

Complete step-by-step answer:

Given equation is

${\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}\left( {{x^2} + x + 1} \right)$

Now consider L.H.S

$ \Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right)$……….. (1)

As we know that ${\cot ^{ - 1}}A + {\tan ^{ - 1}}A = \dfrac{\pi }{2}$.

$ \Rightarrow {\cot ^{ - 1}}A = \dfrac{\pi }{2} - {\tan ^{ - 1}}A$

So, use this property in equation (1) we have,

$ \Rightarrow {\tan ^{ - 1}}x + \dfrac{\pi }{2} - {\tan ^{ - 1}}\left( {x + 1} \right)$

$ = {\tan ^{ - 1}}x - {\tan ^{ - 1}}\left( {x + 1} \right) + \dfrac{\pi }{2}$

Now as we know that ${\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right)$ so, use this property in above equation we have,

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{x - x - 1}}{{1 + x\left( {x + 1} \right)}}} \right) + \dfrac{\pi }{2}$

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{1 + x + {x^2}}}} \right) + \dfrac{\pi }{2}$……….. (2)

Now as we know ${\tan ^{ - 1}}A - {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = \dfrac{\pi }{2}$

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = {\tan ^{ - 1}}A - \dfrac{\pi }{2}$ So, use this property in equation (2) we have,

$ \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}}\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.

Complete step-by-step answer:

Given equation is

${\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}\left( {{x^2} + x + 1} \right)$

Now consider L.H.S

$ \Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}\left( {x + 1} \right)$……….. (1)

As we know that ${\cot ^{ - 1}}A + {\tan ^{ - 1}}A = \dfrac{\pi }{2}$.

$ \Rightarrow {\cot ^{ - 1}}A = \dfrac{\pi }{2} - {\tan ^{ - 1}}A$

So, use this property in equation (1) we have,

$ \Rightarrow {\tan ^{ - 1}}x + \dfrac{\pi }{2} - {\tan ^{ - 1}}\left( {x + 1} \right)$

$ = {\tan ^{ - 1}}x - {\tan ^{ - 1}}\left( {x + 1} \right) + \dfrac{\pi }{2}$

Now as we know that ${\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right)$ so, use this property in above equation we have,

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{x - x - 1}}{{1 + x\left( {x + 1} \right)}}} \right) + \dfrac{\pi }{2}$

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{1 + x + {x^2}}}} \right) + \dfrac{\pi }{2}$……….. (2)

Now as we know ${\tan ^{ - 1}}A - {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = \dfrac{\pi }{2}$

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{A}} \right) = {\tan ^{ - 1}}A - \dfrac{\pi }{2}$ So, use this property in equation (2) we have,

$ \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}}\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.

Last updated date: 25th Sep 2023

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