Question

Prove that \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] where \[xy < 1\]

Answer 1

Hint: In this question, we will proceed by writing the given data and then consider the R.H.S part of the given equation. Then use substitution method along with trigonometry formula to prove that the R.H.S and L.H.S are equal.

Complete step by step answer:
Here we have to prove that \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] where \[xy < 1\]
Now consider the RHS part i.e., \[{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
Let \[x = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}x\] and \[y = \tan \alpha \Rightarrow \alpha = {\tan ^{ - 1}}y\]
So, we have RHS as
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}} = {\tan ^{ - 1}}\left( {\dfrac{{\tan \theta + \tan \alpha }}{{1 - \tan \theta \tan \alpha }}} \right)\]
We know that, \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]. So, using this formula we have
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}} = {\tan ^{ - 1}}\left( {\tan \left( {\theta + \alpha } \right)} \right)\]
Also, we know that \[{\tan ^{ - 1}}\left( {\tan A} \right) = A\]. So, using this formula we have
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}} = \theta + \alpha \]
Resubstituting the values of \[\theta \] and \[\alpha \], we have
\[
   \Rightarrow {\text{R}}{\text{.H}}{\text{.S}} = {\tan ^{ - 1}}x + {\tan ^{ - 1}}y \\
  \therefore {\text{R}}{\text{.H}}{\text{.S}} = {\text{L}}{\text{.H}}{\text{.S}} \\
\]
Hence proved.

Note: Here we have used the trigonometry formulae \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and \[{\tan ^{ - 1}}\left( {\tan A} \right) = A\]. So, for solving these types of problems always remember the formulae in trigonometry and inverse trigonometry.