Answer
453k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)