Question

The value of x satisfying the equation
${\tan ^{ - 1}}x + {\tan ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}(\dfrac{7}{4})$ is equal to
a. $\dfrac{1}{2}$
b. $ - \dfrac{1}{2}$
c. $\dfrac{3}{2}$
d. $\dfrac{1}{3}$

Answer 1

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

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

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