Question

How do you show $\arctan (x) \pm \arctan (y) = \arctan \left[ {\dfrac{{x \pm y}}{{1 \mp xy}}} \right]?$

Answer 1

Hint: First of all consider tangent inverse of $x$ and tangent inverse of $y$ to be some variable. Then from the consideration find the value of $x\;{\text{and}}\;y$ and then use the addition or subtraction formula of tangent with both considered variables. After expanding the addition or subtraction formula of tangent replace the variables and replace the considered variables with original ones.
Addition or subtraction formula of tangent is given as
$\tan (a \pm b) = \dfrac{{\tan a \pm \tan b}}{{1 \mp \tan a\tan b}}$

Formula used:
Addition formula of the tangent function: $\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$
Subtraction formula of the tangent function: $\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$

Complete step by step solution:
To prove the given trigonometric equation $\arctan (x) \pm \arctan (y) = \arctan \left[ {\dfrac{{x \pm y}}{{1 \mp xy}}} \right]$ we will first consider ${\tan ^{ - 1}}x = a\;{\text{and}}\;{\tan ^{ - 1}}y = b$
So we can also write $\tan a = x\;{\text{and}}\;\tan b = y - - - - - (i)$
Now, from the addition or subtraction formula of tangent function, we know that
\[ \Rightarrow \tan (a \pm b) = \dfrac{{\tan a \pm \tan b}}{{1 \mp \tan a\tan b}}\]
Taking both sides inverse tangent function, we will get
\[
   \Rightarrow {\tan ^{ - 1}}\left( {\tan (a \pm b)} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\tan a \pm \tan b}}{{1 \mp \tan a\tan b}}} \right) \\
   \Rightarrow a \pm b = {\tan ^{ - 1}}\left( {\dfrac{{\tan a \pm \tan b}}{{1 \mp \tan a\tan b}}} \right) \\
 \]
Now replacing the considered variables, that is putting ${\tan ^{ - 1}}x = a\;{\text{and}}\;{\tan ^{ - 1}}y = b$ in the equation we will get
\[ \Rightarrow {\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{\tan a \pm \tan b}}{{1 \mp \tan a\tan b}}} \right)\]
And from equation (i), we can further write it as
\[ \Rightarrow {\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x \pm y}}{{1 \mp xy}}} \right)\]
In trigonometry we also denote inverse function of an trigonometric function with prefix “arc” in it, so using this, we can write
\[ \Rightarrow \arctan (x) \pm \arctan (y) = \arctan \left( {\dfrac{{x \pm y}}{{1 \mp xy}}} \right)\]
So we have proven the given trigonometric equation.


Note: The value of domain of “x” and “y” should lie such that their product should not be equals to one, because if their product is equals to one then the argument will become not defined. Also we cannot directly prove this problem so we have considered values first and then proved with help of trigonometric identity.
Domain of inverse tangent function is the set of real numbers whereas its range is given in the interval \[\left[ { - \dfrac{\pi }{2},\;\dfrac{\pi }{2}} \right]\]