# Prove that:

${\text{ta}}{{\text{n}}^{ - 1}}{\text{a}} + {\text{co}}{{\text{t}}^{ - 1}}({\text{a}} + 1) = {\text{ta}}{{\text{n}}^{ - 1}}({{\text{a}}^2} + {\text{a}} + 1).$

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Hint: In this type of question, where we have to prove LHS=RHS for the given inverse trigonometric function. The first step is to convert the expression given in LHS to inverse tangent function if all the expressions are either inverse tangent or inverse cotangent function. Then solve the LHS and it is equal to RHS.

In the given question, we have to prove the given expression. So, the usual approach is to first consider the term given in Left Hand Side(LHS) and simplify it in such a way that it becomes equal to the expression given in Right Hand Side(RHS).

So, we will first consider LHS:

In the question, all the functions are either inverse tangent or inverse cotangent functions. So, we will convert all the functions in LHS to inverse tangent functions.

We know that:

${\text{ta}}{{\text{n}}^{ - 1}}x = {\text{co}}{{\text{t}}^{ - 1}}\dfrac{1}{x};{\text{x > 0}}$

$\therefore $ ${\text{co}}{{\text{t}}^{ - 1}}(a + 1) = {\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{{a + 1}});{\text{a + 1 > 0}}$ .

Therefore, the expression in LHS is written as:

${\text{ta}}{{\text{n}}^{ - 1}}a + {\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{{a + 1}}).$

Also, we know that:

${\text{ta}}{{\text{n}}^{ - 1}}{\text{x}} + {\text{ta}}{{\text{n}}^{ - 1}}{\text{y}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}}} \right);{\text{xy < 1}}$

Therefore, we can write:

\[{\text{ta}}{{\text{n}}^{ - 1}}{\text{a}} + {\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{{{\text{a}} + 1}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{a + }}\dfrac{1}{{{\text{a + 1}}}}}}{{1 - (\dfrac{{\text{a}}}{{{\text{a + 1}}}})}}} \right) = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{{\text{a}}^2} + {\text{a + 1}}}}{{{\text{a + 1 - a}}}}} \right) = {\text{ta}}{{\text{n}}^{ - 1}}\left( {{{\text{a}}^2} + {\text{a}} + 1} \right);\dfrac{{\text{a}}}{{{\text{a + 1}}}}{\text{ < 1}}\]

This expression in LHS is the same as that of the expression in RHS.

Therefore, LHS=RHS proved.

Note: In this type of question, the important step is to change the left hand side expression in terms of a single inverse trigonometric function and then use the inverse trigonometric identities to make the expression in LHS equal to expression in RHS. The condition \[\dfrac{{\text{a}}}{{{\text{a + 1}}}}{\text{ < 1}}\] must be satisfied if we are applying the formula \[{\text{ta}}{{\text{n}}^{ - 1}}{\text{a}} + {\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{{{\text{a}} + 1}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{a + }}\dfrac{1}{{{\text{a + 1}}}}}}{{1 - (\dfrac{{\text{a}}}{{{\text{a + 1}}}})}}} \right)\] .

__Complete step-by-step solution -__In the given question, we have to prove the given expression. So, the usual approach is to first consider the term given in Left Hand Side(LHS) and simplify it in such a way that it becomes equal to the expression given in Right Hand Side(RHS).

So, we will first consider LHS:

In the question, all the functions are either inverse tangent or inverse cotangent functions. So, we will convert all the functions in LHS to inverse tangent functions.

We know that:

${\text{ta}}{{\text{n}}^{ - 1}}x = {\text{co}}{{\text{t}}^{ - 1}}\dfrac{1}{x};{\text{x > 0}}$

$\therefore $ ${\text{co}}{{\text{t}}^{ - 1}}(a + 1) = {\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{{a + 1}});{\text{a + 1 > 0}}$ .

Therefore, the expression in LHS is written as:

${\text{ta}}{{\text{n}}^{ - 1}}a + {\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{{a + 1}}).$

Also, we know that:

${\text{ta}}{{\text{n}}^{ - 1}}{\text{x}} + {\text{ta}}{{\text{n}}^{ - 1}}{\text{y}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}}} \right);{\text{xy < 1}}$

Therefore, we can write:

\[{\text{ta}}{{\text{n}}^{ - 1}}{\text{a}} + {\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{{{\text{a}} + 1}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{a + }}\dfrac{1}{{{\text{a + 1}}}}}}{{1 - (\dfrac{{\text{a}}}{{{\text{a + 1}}}})}}} \right) = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{{\text{a}}^2} + {\text{a + 1}}}}{{{\text{a + 1 - a}}}}} \right) = {\text{ta}}{{\text{n}}^{ - 1}}\left( {{{\text{a}}^2} + {\text{a}} + 1} \right);\dfrac{{\text{a}}}{{{\text{a + 1}}}}{\text{ < 1}}\]

This expression in LHS is the same as that of the expression in RHS.

Therefore, LHS=RHS proved.

Note: In this type of question, the important step is to change the left hand side expression in terms of a single inverse trigonometric function and then use the inverse trigonometric identities to make the expression in LHS equal to expression in RHS. The condition \[\dfrac{{\text{a}}}{{{\text{a + 1}}}}{\text{ < 1}}\] must be satisfied if we are applying the formula \[{\text{ta}}{{\text{n}}^{ - 1}}{\text{a}} + {\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{{{\text{a}} + 1}} = {\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{a + }}\dfrac{1}{{{\text{a + 1}}}}}}{{1 - (\dfrac{{\text{a}}}{{{\text{a + 1}}}})}}} \right)\] .

Last updated date: 03rd Oct 2023

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