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# Evaluate $\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right).$

Last updated date: 18th Sep 2024
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Hint:We know the identity: $\tan \left( {{\theta _1} + {\theta _2}} \right) = \dfrac{{\tan {\theta _1} + \tan {\theta _2}}}{{1 - \tan {\theta _1}\tan {\theta _2}}}$
So by using the above formula we can solve this question.

Complete step by step solution:
Given
$\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right).........................\left( i \right)$

So the in order to solve the above mentioned question we can use identity:
$\tan \left( {{\theta _1} + {\theta _2}} \right) = \dfrac{{\tan {\theta _1} + \tan {\theta _2}}}{{1 - \tan {\theta _1}\tan {\theta _2}}}...............\left( {ii} \right)$

But here all the terms are expressed in tan but in our question it’s expressed in arc tan, such that we have to express the given question in tan and then apply the identity.

Also we know that the tan and arc tan functions are inverse in nature i.e. they are inverse functions:

So converting arc tan to tan, we can write:
$\arctan \left( {\dfrac{1}{2}} \right) \\ \Rightarrow \tan \left( {{\theta _1}} \right) = \dfrac{1}{2}..............\left( {iii} \right) \\ \arctan \left( {\dfrac{1}{3}} \right) \\ \Rightarrow \tan \left( {{\theta _2}} \right) = \dfrac{1}{3}..............\left( {iv} \right) \\$
On observing (iii) and (iv) we know that we have expressed the question in terms of tan such that we can apply directly the identity to find the sum of tangent:
$\Rightarrow \tan \left( {{\theta _1} + {\theta _2}} \right) = \dfrac{{\tan {\theta _1} + \tan {\theta _2}}}{{1 - \tan {\theta _1}\tan {\theta _2}}} \\ \Rightarrow \tan \left( {\left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{3}} \right)} \right) = \dfrac{{\left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{3}} \right)}}{{1 - \left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{1}{3}} \right)}} \\ \\$
Since from (iii) and (iv) we can find the values of $\tan \left( {{\theta _1}} \right)\,{\text{and}}\,\tan \left( {{\theta _2}} \right)$we directly substitute those values in the equation.
$\Rightarrow \tan \left( {\left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{3}} \right)} \right) = \dfrac{{\left( {\dfrac{5}{6}} \right)}}{{1 - \left( {\dfrac{1}{6}} \right)}} = \dfrac{{\left( {\dfrac{5}{6}} \right)}}{{\left( {\dfrac{5}{6}} \right)}} \\ \Rightarrow \tan \left( {\left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{3}} \right)} \right) = 1............\left( v \right) \\$
Now knowing that tan and arc tan functions are inverse in nature i.e. they are inverse functions we can convert (v) back into the terms of arc tan.

Such that:
${\theta _1} + {\theta _2} = \arctan \left( 1 \right)\;\;\;\;{\text{where}}\;{\theta _1} = \arctan \dfrac{1}{2}\;\;{\text{and}}\;\;{\theta _2} = \arctan \dfrac{1}{3}$

Also we know that the value of each individual term${\theta _1}\,\,{\text{and}}\,\,{\theta _2}$ is between $0\,\,{\text{and}}\,\,\dfrac{\pi }{4}$there sum also then should be between
$0\,\,{\text{and}}\,\,\dfrac{\pi }{2}$i.e. in the Quadrant I.

So that we can write:
${\theta _1} + {\theta _2} = \arctan \left( 1 \right) = \dfrac{\pi }{4}..........................\left( {vi} \right)$

Since in the Quadrant I only$\tan \left( {\dfrac{\pi }{4}} \right) = 1$.

Therefore from (vi) we can write our final answer:
$\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$

Note: Some properties useful for solving trigonometric questions:
Quadrant I:$0\; - \;\dfrac{\pi }{2}$ All values are positive.

Quadrant II:$\dfrac{\pi }{2}\; - \;\pi$ Only Sine and Cosec values are positive.

Quadrant III:$\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.

Quadrant IV:$\dfrac{{3\pi }}{2}\; - \;2\pi$ Only Cos and Sec values are positive.

Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side.