3.6 Property 6

2.1 ${\sin ^{ - 1}}x$:
2.2 ${\cos ^{ - 1}}x$:
2.3 ${\tan ^{ - 1}}x$:
2.4 ${\text{cose}}{{\text{c}}^{ - 1}}x$:
2.5 ${\sec^{ - 1}}x$
2.6 ${\cot^{ - 1}}x$
2.7 Summary

3.1 Property 1
3.2 Property 2
3.3 Property 3
3.4 Property 4
3.5 Property 5
3.6 Property 6
3.7 Property 7
3.8 Property 8
3.9 Property 9
3.10 Property 10
3.11 Property 11
3.12 Property 12
3.13 Property 13

(i) ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$\[= \left\{ \begin{gathered} {\tan ^{ - 1}}\frac{{x + y}}{{1 - xy}};xy < 1 \hspace{1em} \\ \pi + {\tan ^{ - 1}}\frac{{x + y}}{{1 - xy}};x > 0,y > 0,xy > 1 \hspace{1em} \\ - \pi + {\tan ^{ - 1}}\frac{{x + y}}{{1 - xy}};x < 0,y < 0,xy > 1 \hspace{1em} \\ \end{gathered} \right\}\]

Proof: Put ${\tan ^{ - 1}}x = A$ and ${\tan ^{ - 1}}y = B$ such that $A,B \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$.

We can write $x = \tan A$ and $y = \tan B$ and apply the formulae of ${\tan \left( {A + B} \right)}$ i.e., $$\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ Put the values of $A$ and $B$, we get $$\tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y} \right) = \frac{{x + y}}{{1 - xy}}$$ Taking inverse both sides we get, $${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)...(1)$$ Now, let us understand why the above proved result is not true for all the values of $x$ and $y$. Let us take the second condition i.e., $${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right);\quad x > 0,y > 0,xy > 1$$ If we consider $x>0$, $y>0$ and $xy>1$, means the denominator of right hand side of equation $(1)$ i.e., $1-xy<0$ and numerator of right hand side of equation $(1)$ is positive. So, we can say that the whole term in right side is negative i.e., $${\frac{{x + y}}{{1 - xy}} < 0}$$ Since we know that the range of ${\tan ^{ - 1}}$ is $\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$.

From the graph of function $y = {\tan ^{ - 1}}x$, we can say that if $x>0$, $y \in \left( {0,\frac{\pi }{2}} \right)$ and if $x<0$, $y \in \left( { - \frac{\pi }{2},0} \right)$. Using this concept we can say that in the equation $(1)$, range of right hand side is $\left( { - \frac{\pi }{2},0} \right)$ as the term in ${\tan ^{ - 1}}$ is negative and the range of left hand side is $\left( {0,\frac{\pi }{2}} \right)$ for ${\tan ^{ - 1}}x$ and $\left( {0,\frac{\pi }{2}} \right)$ for ${\tan ^{ - 1}}y$ which gives us the combined range as $\left( {0,\pi } \right)$.

In order to make the range equal on both the sides of the equation, we add $\pi$ in right hand side and the equation becomes $${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}} < 0} \right);\quad x > 0,y > 0,xy > 1$$

Using the same concept, we subtract $\pi$ in the equation to get the third condition i.e., $${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = - \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}} < 0} \right);\quad x < 0,y < 0,xy > 1$$

(ii) ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y$\[= \left\{ \begin{gathered} {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right);xy > - 1 \hspace{1em} \\ \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right);x > 0,y < 0,xy < - 1 \hspace{1em} \\ - \pi + {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right);x < 0,y > 0,xy < - 1 \hspace{1em} \\ \end{gathered} \right\}\]