Question

Solve \[{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \dfrac{\pi }{4}\].

Answer 1

Hint:In the given question, we have an inverse trigonometric function , therefore can’t apply a direct formula for \[\tan x + \tan y\] . For inverse trigonometric function we have conditions , and based on that conditions we have different formulas , like if \[xy < 1\] then we will use \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] and if \[xy > 1\] , then we will use \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\].

Complete step by step answer:
Given : \[{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \dfrac{\pi }{4}\]
Now given the equation the value of \[xy\] is greater than \[1\] . Since , \[6 > 1\] .
Therefore we will use formula \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\] .
Using the formula we get ,
\[\pi + {\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - \left( {2x} \right)\left( {3x} \right)}}} \right) = \dfrac{\pi }{4}\]
On simplifying we get ,
\[{\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - 6{x^2}}}} \right) = \dfrac{\pi }{4} - \pi \]
On solving we get ,
\[{\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - 6{x^2}}}} \right) = \dfrac{{ - 3\pi }}{4}\]
On simplifying we get ,
\[\dfrac{{2x + 3x}}{{1 - 6{x^2}}} = \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)\]

Now putting the value of \[\tan \left( {\dfrac{{ - 3\pi }}{4}} \right)\] as \[ - 1\] , we get
\[\dfrac{{2x + 3x}}{{1 - 6{x^2}}} = - 1\]
On simplifying we get ,
\[5x = - 1\left( {1 - 6{x^2}} \right)\]
On solving we get ,
\[5x = - 1 + 6{x^2}\]
On simplifying we get ,
\[6{x^2} - 5x - 1 = 0\]
Now we will factorize the above quadratic polynomial to get the required answer , therefore we can write the equation as :
\[6{x^2} - 6x - x - 1 = 0\]
On solving we get ,
\[6x\left( {x - 1} \right) + 1\left( {x - 1} \right) = 0\]
Now taking \[\left( {x - 1} \right)\] common we get ,
\[\left( {6x + 1} \right)\left( {x - 1} \right) = 0\]
Now equating both terms with zero we get ,
\[\therefore x = \dfrac{{ - 1}}{6},1\] .

Therefore , the required solution for the given inverse trigonometric function is \[ = \dfrac{{ - 1}}{6},1\].

Note: In the given questions never use the direct formula of \[\tan x + \tan y\] in inverse trigonometric function as inverse function formulas applied on some fixed conditions . Always check the conditions for \[xy < 1\] and \[xy > 1\] and then apply the formulas . The inverse trigonometric functions perform the opposite operations of the trigonometric functions such as sine , cosine , tangent , cosecant , secant and cotangent . Inverse Trigonometric Functions are defined in a certain period (under certain domains) .