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

Solve tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \dfrac{\pi }{4}.

Explanation

Solution

In the given question, we have an inverse trigonometric function , therefore can’t apply a direct formula for tan⁡x+tan⁡y\tan x + \tan y . For inverse trigonometric function we have conditions , and based on that conditions we have different formulas , like if xy<1xy < 1 then we will use tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) and if xy>1xy > 1 , then we will use tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right).

Complete step by step answer:
Given : tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \dfrac{\pi }{4}
Now given the equation the value of xyxy is greater than 11 . Since , 6>16 > 1 .
Therefore we will use formula tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) .
Using the formula we get ,
π+tan⁡−1(2x+3x1−(2x)(3x))=π4\pi + {\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - \left( {2x} \right)\left( {3x} \right)}}} \right) = \dfrac{\pi }{4}
On simplifying we get ,
tan⁡−1(2x+3x1−6x2)=π4−π{\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - 6{x^2}}}} \right) = \dfrac{\pi }{4} - \pi
On solving we get ,
tan⁡−1(2x+3x1−6x2)=−3π4{\tan ^{ - 1}}\left( {\dfrac{{2x + 3x}}{{1 - 6{x^2}}}} \right) = \dfrac{{ - 3\pi }}{4}
On simplifying we get ,
2x+3x1−6x2=tan⁡(−3π4)\dfrac{{2x + 3x}}{{1 - 6{x^2}}} = \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)

Now putting the value of tan⁡(−3π4)\tan \left( {\dfrac{{ - 3\pi }}{4}} \right) as −1 - 1 , we get
2x+3x1−6x2=−1\dfrac{{2x + 3x}}{{1 - 6{x^2}}} = - 1
On simplifying we get ,
5x=−1(1−6x2)5x = - 1\left( {1 - 6{x^2}} \right)
On solving we get ,
5x=−1+6x25x = - 1 + 6{x^2}
On simplifying we get ,
6x2−5x−1=06{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 :
6x2−6x−x−1=06{x^2} - 6x - x - 1 = 0
On solving we get ,
6x(x−1)+1(x−1)=06x\left( {x - 1} \right) + 1\left( {x - 1} \right) = 0
Now taking (x−1)\left( {x - 1} \right) common we get ,
(6x+1)(x−1)=0\left( {6x + 1} \right)\left( {x - 1} \right) = 0
Now equating both terms with zero we get ,
∴x=−16,1\therefore x = \dfrac{{ - 1}}{6},1 .

Therefore , the required solution for the given inverse trigonometric function is =−16,1 = \dfrac{{ - 1}}{6},1.

Note: In the given questions never use the direct formula of tan⁡x+tan⁡y\tan x + \tan y in inverse trigonometric function as inverse function formulas applied on some fixed conditions . Always check the conditions for xy<1xy < 1 and xy>1xy > 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) .