Question

If \[{\tan ^{ - 1}}\left\{ {\dfrac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}} \right\} = \alpha \], then ${x^2}$$ = $
$
  A.\sin 2\alpha \\
  B.\sin \alpha \\
  C.\cos 2\alpha \\
  D.\cos \alpha \\
$

Answer 1

Hint:In this question we will use the concept of componendo and dividendo i.e. \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\] and some basis conversions like $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ and $2\sin \alpha \cos \alpha = \sin 2\alpha $. Use this to find the value of ${x^2}$.

Complete step-by-step answer:
According to the question it is given \[{\tan ^{ - 1}}\left\{ {\dfrac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}} \right\} = \alpha \],
Applying tan on both side, we can write it as
\[\dfrac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }} = \dfrac{{\tan \alpha }}{1}\] $\left[ {\dfrac{{\tan \alpha }}{1}} \right.$ because componendo and dividendo is used i.e.$\left. {\dfrac{a}{b} = \dfrac{c}{d}} \right]$
 We know that in componendo and dividendo \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\]
Now,
$
   \Rightarrow \dfrac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }} = \dfrac{{\tan \alpha + 1}}{{\tan \alpha - 1}} \\
   \Rightarrow \dfrac{{2\sqrt {1 + {x^2}} }}{{ - 2\sqrt {1 - {x^2}} }} = \dfrac{{\tan \alpha + 1}}{{\tan \alpha - 1}} \\
$
Change the sign so we can write,
$ \Rightarrow \dfrac{{\sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} }} = \dfrac{{1 + \tan \alpha }}{{1 - \tan \alpha }}$
Now we can write $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
Hence,
$
   \Rightarrow \dfrac{{\sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} }} = \dfrac{{1 + \dfrac{{\sin \alpha }}{{\cos \alpha }}}}{{1 - \dfrac{{\sin \alpha }}{{\cos \alpha }}}} \\
   \Rightarrow \dfrac{{\sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} }} = \dfrac{{\dfrac{{\cos \alpha + \sin \alpha }}{{\cos \alpha }}}}{{\dfrac{{\cos \alpha - \sin \alpha }}{{\cos \alpha }}}} \\
$
Squaring on both the sides,
\[
   \Rightarrow \dfrac{{1 + {x^2}}}{{1 - {x^2}}} = \dfrac{{{{\left( {\cos \alpha + \sin \alpha } \right)}^2}}}{{{{\left( {\cos \alpha - \sin \alpha } \right)}^2}}} \\
   \Rightarrow \dfrac{{1 + {x^2}}}{{1 - {x^2}}} = \dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha + 2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha + {{\sin }^2}\alpha - 2\sin \alpha \cos \alpha }} \\
   \Rightarrow \dfrac{{1 + {x^2}}}{{1 - {x^2}}} = \dfrac{{1 + 2\sin \alpha \cos \alpha }}{{1 - 2\sin \alpha \cos \alpha }} \\
 \]
We know that $2\sin \alpha \cos \alpha = \sin 2\alpha $
Hence, by substituting the value we get
$ \Rightarrow \dfrac{{1 + {x^2}}}{{1 - {x^2}}} = \dfrac{{1 + \sin 2\alpha }}{{1 - \sin 2\alpha }}$
Apply componendo and dividendo again, we get
$
   \Rightarrow \dfrac{{1 + {x^2} + 1 - {x^2}}}{{1 + {x^2} - 1 + {x^2}}} = \dfrac{{1 + \sin 2\alpha + 1 - \sin 2\alpha }}{{1 + \sin 2\alpha - 1 + \sin 2\alpha }} \\
   \Rightarrow \dfrac{2}{{2{x^2}}} = \dfrac{2}{{2\sin 2\alpha }} \\
   \Rightarrow {x^2} = \sin 2\alpha \\
 $
So the value of ${x^2}$ is $\sin 2\alpha$

Note: It is always advisable to remember some basic concepts and basic conversions while involving trigonometric questions as it saves a lot of time.For above question we used concept of componendo and dividendo i.e. \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\] for solving the problem.Students make mistakes while applying componendo and dividendo have to take care of values which is a and b from the given equation and to find a+b and a-b values and also should remember trigonometric ratios and identities for solving these types of problems.