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Question: Prove that \[{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2...

Prove that tan⁡−1[1+x2+1−x21+x2−1−x2]{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]=π4+12cos⁡−1(x2) = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)

Explanation

Solution

Hint : Use trigonometric substitution. Substitute xx in such a way that 1+x21 + {x^2} becomes square of some term. Then it would come out of square root sign and you will be able to solve the question.

Complete step-by-step answer :
L.H.S =tan⁡−1[1+x2+1−x21+x2−1−x2] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]
Put x2=cos⁡2θ{x^2} = \cos 2\theta
⇒tan⁡−1[1+x2+1−x21+x2−1−x2]=tan⁡−1[1+cos⁡2θ+1−cos⁡2θ1+cos⁡2θ−1−cos⁡2θ]\Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right] . . . (1)
We know that,
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
=cos⁡2θ−(1−cos⁡2θ)= {\cos ^2}\theta - (1 - {\cos ^2}\theta )
=cos⁡2θ−1+cos⁡2θ= {\cos ^2}\theta - 1 + {\cos ^2}\theta
⇒cos⁡2θ=2cos⁡2θ−1\Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1
Substitute this value in equation (1). We get
tan⁡−1[1+cos⁡2θ+1−cos⁡2θ1+cos⁡2θ−1−cos⁡2θ]=tan⁡−1[1+2cos⁡2θ−1+1−(2cos⁡2θ−1)1+2cos⁡2θ−1−1−(2cos⁡2θ−1)]{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + 2{{\cos }^2}\theta - 1} + \sqrt {1 - (2{{\cos }^2}\theta - 1)} }}{{\sqrt {1 + 2{{\cos }^2}\theta - 1} - \sqrt {1 - (2{{\cos }^2}\theta - 1)} }}} \right]
By simplifying it, we get
=tan⁡−1[2cos⁡2θ+1−2cos⁡2θ+12cos⁡2θ−1−2cos⁡2θ+1]= {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {1 - 2{{\cos }^2}\theta + 1} }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {1 - 2{{\cos }^2}\theta + 1} }}} \right]
=tan⁡−1[2cos⁡2θ+2−2cos⁡2θ2cos⁡2θ−2−2cos⁡2θ]= {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2 - 2{{\cos }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2 - 2{{\cos }^2}\theta } }}} \right]
=tan⁡−1[2cos⁡2θ+2(1−cos⁡2θ)2cos⁡2θ−2(1−cos⁡2θ)]= {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2(1 - {{\cos }^2}\theta } )}}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2(1 - {{\cos }^2}\theta )} }}} \right]
=tan⁡−1[2cos⁡2θ+2sin⁡2θ2cos⁡2θ−2sin⁡2θ]= {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right]
Dividing numerator and denominator by 2\sqrt 2 inside tan⁡−1{\tan ^{ - 1}}, we get
tan⁡−1[2cos⁡2θ+2sin⁡2θ2cos⁡2θ−2sin⁡2θ]=tan⁡−1[cos⁡2θ+sin⁡2θcos⁡2θ−sin⁡2θ]{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {{{\cos }^2}\theta } + \sqrt {{{\sin }^2}\theta } }}{{\sqrt {{{\cos }^2}\theta } - \sqrt {{{\sin }^2}\theta } }}} \right]
=tan⁡−1[cos⁡θ+sin⁡θcos⁡θ−sin⁡θ]= {\tan ^{ - 1}}\left[ {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right]
Dividing numerator and denominator by cos⁡θ\cos \theta inside tan⁡−1{\tan ^{ - 1}}, we get
tan⁡−1[cos⁡θ+sin⁡θcos⁡θ−sin⁡θ]=tan⁡−1[1+sin⁡θcos⁡θ1−sin⁡θcos⁡θ]{\tan ^{ - 1}}\left[ {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{1 + \dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}} \right]
=tan⁡−1[1+tan⁡θ1−tan⁡θ]= {\tan ^{ - 1}}\left[ {\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right]
=tan⁡−1[1+tan⁡θ1−1×tan⁡θ]= {\tan ^{ - 1}}\left[ {\dfrac{{1 + \tan \theta }}{{1 - 1 \times \tan \theta }}} \right] (∵1×tan⁡θ=tan⁡θ)\left( {\because 1 \times \tan \theta = \tan \theta } \right)
=tan⁡−1[tan⁡(π4)+tan⁡θ1−tan⁡(π4)×tan⁡θ]= {\tan ^{ - 1}}\left[ {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan \theta }}{{1 - \tan \left( {\dfrac{\pi }{4}} \right) \times \tan \theta }}} \right] (∵tan⁡(π4)=1)\left( {\because \tan \left( {\dfrac{\pi }{4}} \right) = 1} \right)
=tan⁡−1(tan⁡(π4+θ))= {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} + \theta } \right)} \right) (∵tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B)\left( {\because \tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right)
=π4+θ= \dfrac{\pi }{4} + \theta (∵tan⁡−1(tan⁡θ)=θ)\left( {\because {{\tan }^{ - 1}}(\tan \theta ) = \theta } \right) . . . (2)
Now,
x2=cos⁡2θ{x^2} = \cos 2\theta
⇒cos⁡−1(x2)=2θ\Rightarrow {\cos ^{ - 1}}\left( {{x^2}} \right) = 2\theta
By rearranging it, we get
2θ=cos⁡−1(x2)2\theta = {\cos ^{ - 1}}\left( {{x^2}} \right)
⇒θ=12cos⁡−1(x2)\Rightarrow \theta = \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)
Substituting this value in equation (2), we get
π4+θ=π4+12cos⁡−1(x2)\dfrac{\pi }{4} + \theta = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)
= R.H.S.
Hence, it is proved that
tan⁡−1[1+x2+1−x21+x2−1−x2]=π4+12cos⁡−1(x2){\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right] = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)

Note : For the question to exist, 1−x2\sqrt {1 - {x^2}} should exist.
For 1−x2\sqrt {1 - {x^2}} to exist, 1−x2⩾01 - {x^2} \geqslant 0
Because, the square root of a negative term will not exist.
⇒1⩾x2\Rightarrow 1 \geqslant {x^2}
⇒x2⩽1\Rightarrow {x^2} \leqslant 1 . . . (2)
But a square term is never negative.
⇒x2⩾0\Rightarrow {x^2} \geqslant 0 . . . (3)
From equation (2) and (3), we get
0⩽x2⩽10 \leqslant {x^2} \leqslant 1
⇒0⩽cos⁡2θ⩽1\Rightarrow 0 \leqslant \cos 2\theta \leqslant 1 (∵x2=cos⁡2θ)\left( {\because {x^2} = \cos 2\theta } \right)
Therefore, we could substitute, x2=cos⁡2θ{x^2} = \cos 2\theta