Question

Prove that the trigonometric expression $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ holds true.

Answer 1

Hint – In this question consider the L.H.S part and break $2{\tan ^{ - 1}}x$ into ${\tan ^{ - 1}}x + {\tan ^{ - 1}}x$, then apply the formula${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)$. Then convert trigonometric identity of ${\tan ^{ - 1}}x{\text{ into si}}{{\text{n}}^{ - 1}}x$ to get the proof.

Complete step-by-step solution -

Proof –
Consider L.H.S
$ \Rightarrow 2{\tan ^{ - 1}}x$
This is written as
$ \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}x$
Now as we know that ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)$ so use this property in above equation we have,
$ \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{x + x}}{{1 - {x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$
Now let
y = ${\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$................. (1)
$ \Rightarrow \tan y = \left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$
Now as we know tan is ratio of perpendicular to base
Therefore perpendicular = 2x
And base = $(1 – {x^2})$.
Now apply Pythagoras theorem in right triangle ABC as shown above we have,
\[ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}\]
\[ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {2x} \right)^2} + {\left( {1 - {x^2}} \right)^2}\]
Now simplify this we have,
\[ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = 4{x^2} + 1 + {x^4} - 2{x^2} = 1 + 2{x^2} + {x^4} = {\left( {1 + {x^2}} \right)^2}\]
Therefore, \[\left( {{\text{Hypotenuse}}} \right) = \left( {1 + {x^2}} \right)\]
Now as we know sin is the ratio of perpendicular to hypotenuse so we have,
$ \Rightarrow \sin y = \dfrac{{2x}}{{1 + {x^2}}}$
$ \Rightarrow y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$
So from equation (1) we have,
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$
= L.H.S
Hence proved.

Note – The conversion of one inverse trigonometric identity into another is based upon the same concept as that of conversion of a normal trigonometric ratio into another. A right angled triangle depicting sides of perpendicular, base and hypotenuse, helps establish relations.