Question

The value of \[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\]
A) 1
B) Not defined
C) \[\sqrt 3 \]
D) \[\dfrac{1}{{\sqrt 3 }}\]

Answer 1

Hint: Here we will first convert \[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}\] into the terms of \[{\tan ^{ - 1}}\theta \] and then use the formula of \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B\] to get the desired solution.

Formula used:
\[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
\[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\]

Complete step-by-step answer:
The given expression is: -
\[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\]
We will first convert \[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}\] into the terms of \[{\tan ^{ - 1}}\theta \]
We know that:
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x\]
Also,
\[2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}\]
Hence,
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}\]
Therefore, the expression becomes:
\[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\tan }^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}} \right)\]
Now applying the following formula:
\[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
We get:-
\[ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{{1 - {x^2}}}{{2x}} + \dfrac{{2x}}{{1 - {x^2}}}}}{{1 - \left( {\dfrac{{1 - {x^2}}}{{2x}}} \right)\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)}}} \right)\]
Simplifying it further we get:-
\[ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{{1 - {x^2}}}{{2x}} + \dfrac{{2x}}{{1 - {x^2}}}}}{{1 - 1}}} \right)\]
Taking the LCM and solving it further we get:-
\[ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{\left( {1 - {x^2}} \right)\left( {1 - {x^2}} \right) + 2x\left( {2x} \right)}}{{0\left[ {\left( {2x} \right)\left( {1 - {x^2}} \right)} \right]}}} \right)\]
Simplifying it further we get:-
\[ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{\left( {1 - {x^2}} \right)\left( {1 - {x^2}} \right) + 2x\left( {2x} \right)}}{0}} \right)\]
Now we know that:
\[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\]
Hence applying this formula we get:-
\[ = \dfrac{{\left( {1 - {x^2}} \right)\left( {1 - {x^2}} \right) + 2x\left( {2x} \right)}}{0}\]
Now we know that:
Anything which is divided by 0 is not defined
Hence, we get:-
\[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\tan }^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}} \right) = {\text{not defined}}\]
This implies,
\[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = {\text{not defined}}\]

Therefore, option B is the correct option.

Note: Students can also convert \[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}\] into the terms of \[{\tan ^{ - 1}}\theta \] using the following formula:-
\[{\cos ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right)\]
Therefore,
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {{\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)}^2}} }}{{\dfrac{{1 - {x^2}}}{{1 + {x^2}}}}}} \right)\]
Solving it further we get:-
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sqrt {{{\left( {1 + {x^2}} \right)}^2} - {{\left( {1 - {x^2}} \right)}^2}} }}{{1 + {x^2}}}}}{{\dfrac{{1 - {x^2}}}{{1 + {x^2}}}}}} \right)\]
Cancelling the terms we get:-
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {{{\left( {1 + {x^2}} \right)}^2} - {{\left( {1 - {x^2}} \right)}^2}} }}{{1 - {x^2}}}} \right)\]
Now using the following identity:-
\[{\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab\]
We get:-
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {4\left( 1 \right)\left( {{x^2}} \right)} }}{{1 - {x^2}}}} \right)\]
Solving it further we get:-
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {4{x^2}} }}{{1 - {x^2}}}} \right)\]
Taking the square root we get:-
\[{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]