Question

Prove that \[{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]\]$ = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)$

Answer 1

Hint: Use trigonometric substitution. Substitute $x$ in such a way that $1 + {x^2}$ 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 \[ = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]\]
Put ${x^2} = \cos 2\theta $
$ \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 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
$ = {\cos ^2}\theta - (1 - {\cos ^2}\theta )$
$ = {\cos ^2}\theta - 1 + {\cos ^2}\theta $
$ \Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1$
Substitute this value in equation (1). We get
\[{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + 2{{\cos }^2}\theta - 1} + \sqrt {1 - (2{{\cos }^2}\theta - 1)} }}{{\sqrt {1 + 2{{\cos }^2}\theta - 1} - \sqrt {1 - (2{{\cos }^2}\theta - 1)} }}} \right]\]
By simplifying it, we get
\[ = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {1 - 2{{\cos }^2}\theta + 1} }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {1 - 2{{\cos }^2}\theta + 1} }}} \right]\]
\[ = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2 - 2{{\cos }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2 - 2{{\cos }^2}\theta } }}} \right]\]
\[ = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2(1 - {{\cos }^2}\theta } )}}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2(1 - {{\cos }^2}\theta )} }}} \right]\]
\[ = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right]\]
Dividing numerator and denominator by $\sqrt 2 $ inside ${\tan ^{ - 1}}$, we get
\[{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {{{\cos }^2}\theta } + \sqrt {{{\sin }^2}\theta } }}{{\sqrt {{{\cos }^2}\theta } - \sqrt {{{\sin }^2}\theta } }}} \right]\]
\[ = {\tan ^{ - 1}}\left[ {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right]\]
Dividing numerator and denominator by \[\cos \theta \] inside ${\tan ^{ - 1}}$, we get
\[{\tan ^{ - 1}}\left[ {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right] = {\tan ^{ - 1}}\left[ {\dfrac{{1 + \dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}} \right]\]
$ = {\tan ^{ - 1}}\left[ {\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right]$
$ = {\tan ^{ - 1}}\left[ {\dfrac{{1 + \tan \theta }}{{1 - 1 \times \tan \theta }}} \right]$ $\left( {\because 1 \times \tan \theta = \tan \theta } \right)$
$ = {\tan ^{ - 1}}\left[ {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan \theta }}{{1 - \tan \left( {\dfrac{\pi }{4}} \right) \times \tan \theta }}} \right]$ $\left( {\because \tan \left( {\dfrac{\pi }{4}} \right) = 1} \right)$
$ = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} + \theta } \right)} \right)$ $\left( {\because \tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right)$
$ = \dfrac{\pi }{4} + \theta $ $\left( {\because {{\tan }^{ - 1}}(\tan \theta ) = \theta } \right)$ . . . (2)
Now,
${x^2} = \cos 2\theta $
$ \Rightarrow {\cos ^{ - 1}}\left( {{x^2}} \right) = 2\theta $
By rearranging it, we get
$2\theta = {\cos ^{ - 1}}\left( {{x^2}} \right)$
$ \Rightarrow \theta = \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)$
Substituting this value in equation (2), we get
$\dfrac{\pi }{4} + \theta = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)$
= R.H.S.
Hence, it is proved that
\[{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right] = \dfrac{\pi }{4} + \dfrac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)\]

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