Question

How do you find the value of $\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$?

Answer 1

Hint: To find the value of given trigonometric function we will use trigonometric identity ${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$. After implementing this identity we will find the value of the $\sin $ function.

Complete step by step answer:
Here we are given to simplify the term $\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$
First, we solve $\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$
We know that
${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$
Putting $x = \dfrac{1}{2}$ and $y = \dfrac{1}{3}$ in the above formula. We get,
$ \Rightarrow $$\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2} \times \dfrac{1}{3}}}} \right)$
Simplifying the above equation. We get,
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{3 + 2}}{6}}}{{1 - \dfrac{1}{6}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{\dfrac{{6 - 1}}{6}}}} \right)$
Further solving the above equation. We get,
$ \Rightarrow $${\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}}} \right)$$ = {\tan ^{ - 1}}\left( {\dfrac{5}{6} \times \dfrac{6}{5}} \right)$
$ \Rightarrow {\tan ^{ - 1}}(1)$
Let ${\tan ^{ - 1}}(1) = A$
So, $\tan (A) = 1$
We know that $\tan (45^\circ ) = 1$

Here, we get $A = 45^\circ $
Therefore, ${\tan ^{ - 1}}(1) = 45^\circ $
So, $\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$$ = 45^\circ $
Therefore,
$ \Rightarrow $$\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$$ = \sin (45^\circ )$
We know that $\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }}$
Therefore, we get
$\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = \dfrac{1}{{\sqrt 2 }}$
Hence, the value of $\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ is equal to $\dfrac{1}{{\sqrt 2 }}$.

Note: In this types of problems the student must remember the basic trigonometric identities and the value of angles of trigonometric function such as $\tan (45^\circ ) = 1$ and $\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }}$.
In these types of problems first reduce the function by using different trigonometric identities and then solve it. After reducing it will be easier to solve the problem.