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Question: The value of x satisfying the equation \({\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{2}{3}) = {\tan ^...

The value of x satisfying the equation
tan⁡−1x+tan⁡−1(23)=tan⁡−1(74){\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}(\dfrac{7}{4}) is equal to
a. 12\dfrac{1}{2}
b. −12 - \dfrac{1}{2}
c. 32\dfrac{3}{2}
d. 13\dfrac{1}{3}

Explanation

Solution

Hint: Here, we will find the value of x by using the inverse trigonometric formulaetan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right).

Complete step-by-step answer:
We have to find the value of x satisfying tan⁡−1x+tan⁡−1(23)=tan⁡−1(74){\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}(\dfrac{7}{4}) using the formulae-
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right).
⇒\Rightarrow tan⁡−1(x+231−23x)=tan⁡−1(74){\tan ^{ - 1}}\left( {\dfrac{{x + \dfrac{2}{3}}}{{1 - \dfrac{2}{3}x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{7}{4}} \right)
Taking LCM in the left hand side we have
⇒\Rightarrow tan⁡−1(3x+233−2x3)=tan⁡−1(74){\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{3x + 2}}{3}}}{{\dfrac{{3 - 2x}}{3}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{7}{4}} \right)
On comparing both sides, we have
⇒\Rightarrow 3x+233−2x3=74\dfrac{{\dfrac{{3x + 2}}{3}}}{{\dfrac{{3 - 2x}}{3}}} = \dfrac{7}{4}
⇒\Rightarrow 4(3x+2)=7(3−2x)4(3x + 2) = 7(3 - 2x)
⇒\Rightarrow 12x+14x=21−812x+14x=21-8
Hence, we get x=1326=12x = \dfrac{{13}}{{26}} = \dfrac{1}{2}
So option (a) is the right answer.

Note: Whenever we come across such problems, we need to use the formula for tan⁡−1{\tan ^{ - 1}} addition or subtraction of two quantities for simplification.