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Question: The value of \[\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - ...

The value of tan⁡(tan⁡−11−x22x+cos⁡−11−x21+x2)\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) 3\sqrt 3
D) 13\dfrac{1}{{\sqrt 3 }}

Explanation

Solution

Here we will first convert cos⁡−11−x21+x2{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} into the terms of tan⁡−1θ{\tan ^{ - 1}}\theta and then use the formula of tan⁡−1A+tan⁡−1B{\tan ^{ - 1}}A + {\tan ^{ - 1}}B to get the desired solution.

Formula used:
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
tan⁡(tan⁡−1x)=x\tan \left( {{{\tan }^{ - 1}}x} \right) = x

Complete step-by-step answer:
The given expression is: -
tan⁡(tan⁡−11−x22x+cos⁡−11−x21+x2)\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)
We will first convert cos⁡−11−x21+x2{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} into the terms of tan⁡−1θ{\tan ^{ - 1}}\theta
We know that:
cos⁡−11−x21+x2=2tan⁡−1x{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x
Also,
2tan⁡−1x=tan⁡−12x1−x22{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}
Hence,
cos⁡−11−x21+x2=tan⁡−12x1−x2{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}
Therefore, the expression becomes:
tan⁡(tan⁡−11−x22x+tan⁡−12x1−x2)\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\tan }^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}} \right)
Now applying the following formula:
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
We get:-
=tan⁡(tan⁡−11−x22x+2x1−x21−(1−x22x)(2x1−x2))= \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⁡(tan⁡−11−x22x+2x1−x21−1)= \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⁡(tan⁡−1(1−x2)(1−x2)+2x(2x)0[(2x)(1−x2)])= \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⁡(tan⁡−1(1−x2)(1−x2)+2x(2x)0)= \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⁡(tan⁡−1x)=x\tan \left( {{{\tan }^{ - 1}}x} \right) = x
Hence applying this formula we get:-
=(1−x2)(1−x2)+2x(2x)0= \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⁡(tan⁡−11−x22x+tan⁡−12x1−x2)=not defined\tan \left( {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\tan }^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}} \right) = {\text{not defined}}
This implies,
tan⁡(tan⁡−11−x22x+cos⁡−11−x21+x2)=not defined\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⁡−11−x21+x2{\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} into the terms of tan⁡−1θ{\tan ^{ - 1}}\theta using the following formula:-
cos⁡−1x=tan⁡−1(1−x2x){\cos ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right)
Therefore,
cos⁡−11−x21+x2=tan⁡−1(1−(1−x21+x2)21−x21+x2){\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⁡−11−x21+x2=tan⁡−1((1+x2)2−(1−x2)21+x21−x21+x2){\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⁡−11−x21+x2=tan⁡−1((1+x2)2−(1−x2)21−x2){\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:-
(a+b)2−(a−b)2=4ab{\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab
We get:-
cos⁡−11−x21+x2=tan⁡−1(4(1)(x2)1−x2){\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⁡−11−x21+x2=tan⁡−1(4x21−x2){\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⁡−11−x21+x2=tan⁡−1(2x1−x2){\cos ^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}} = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)