Answer

Verified

406.5k+ views

**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:**

$\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$

$\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Onam is the main festival of which state A Karnataka class 7 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Who was the founder of muslim league A Mohmmad ali class 10 social science CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers