Question

If ${\tan ^{ - 1}}\left( {\dfrac{{x - 1}}{{x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{2x - 1}}{{2x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{23}}{{36}}} \right)$ then $x =$
$A)\dfrac{3}{4},\dfrac{{ - 3}}{8}$
$B)\dfrac{3}{4},\dfrac{3}{8}$
$C)\dfrac{4}{3},\dfrac{3}{8}$
$D)$ None of these

Answer 1

The given question deals with simplification of trigonometric identities. Whenever this type of question is given, the very first step should be if the question can be simplified using any standard trigonometric identities. Here, we can easily notice that the standard formula , i.e. ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$ can be applied for simplification. So basic algebraic rules and identities have to be kept in mind while solving these types of questions.

Complete step by step answer:
Given,
${\tan ^{ - 1}}\left( {\dfrac{{x - 1}}{{x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{2x - 1}}{{2x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{23}}{{36}}} \right)$ $......(1)$
By Identity;
$\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$ $......(2)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ , we know;
$\Rightarrow x = \left( {\dfrac{{x - 1}}{{x + 1}}} \right)$ and $y = \left( {\dfrac{{2x - 1}}{{2x + 1}}} \right)$
Using the trigonometric identity given in equation $\left( 2 \right)$, we get;
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{x - 1}}{{x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{2x - 1}}{{2x + 1}}} \right) = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 1}}{{x + 1}} + \dfrac{{2x - 1}}{{2x + 1}}}}{{1 - \left( {\dfrac{{x - 1}}{{x + 1}}} \right) \times \left( {\dfrac{{2x - 1}}{{2x + 1}}} \right)}}} \right]$
According to the given question;
$\Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 1}}{{x + 1}} + \dfrac{{2x - 1}}{{2x + 1}}}}{{1 - \left( {\dfrac{{x - 1}}{{x + 1}}} \right) \times \left( {\dfrac{{2x - 1}}{{2x + 1}}} \right)}}} \right] = {\tan ^{ - 1}}\left( {\dfrac{{23}}{{46}}} \right)$
The inverse function on L.H.S. and R.H.S. gets nullified, so our expression gets reduced to;
$= \left[ {\dfrac{{\dfrac{{\left( {2x + 1} \right)\left( {x - 1} \right) + \left( {2x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}}}}{{\dfrac{{\left( {x + 1} \right)\left( {2x + 1} \right) - \left( {x - 1} \right)\left( {2x - 1} \right)}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}}}}} \right]$
On further simplification;
$= \dfrac{{2{x^2} - 2x + x - 1 + 2{x^2} + 2x - x - 1}}{{2{x^2} + x + 2x + 1 - \left( {2{x^2} - x - 2x + 1} \right)}} = \dfrac{{23}}{{46}}$
$\Rightarrow \dfrac{{4{x^2} - 2}}{{6x}} = \dfrac{{23}}{{46}}$
Simplifying ;
$\Rightarrow \dfrac{{4{x^2} - 2}}{x} = \dfrac{{23}}{6}$
Rearranging the above equation;
$\Rightarrow \left( {4{x^2} - 2} \right) \times 6 = 23x$
$\Rightarrow 24{x^2} - 12 = 23x$
Taking all the terms to L.H.S.
$\Rightarrow 24{x^2} - 23x - 12 = 0$ $.......\left( 3 \right)$
Now, we will solve for the value of x;
By quadratic equation formula for $a{x^2} + bx + c = 0$ , the value of $x$ can be calculated as;$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
By equation $\left( 3 \right)$ , we know $a = 24$ , $b = - 23$ and $c = - 12$ ; Putting the values in the quadratic formula, we get;
$\Rightarrow x = \dfrac{{23 + \sqrt {{{\left( {23} \right)}^2} - 4 \times 24 \times \left( { - 12} \right)} }}{{2 \times 24}}$
$\Rightarrow x = \dfrac{{23 \pm 41}}{{48}}$
The above equation gives us two values of $x$;
$\Rightarrow x = \dfrac{4}{3}$ and $x = - \dfrac{3}{8}$
Therefore, the values of $x$ are $\dfrac{4}{3}$ and $- \dfrac{3}{8}$.
The value of $x$ can also be found out by factorizing the equation;
$\Rightarrow 24{x^2} - 23x - 12 = 0$
$\Rightarrow 24{x^2} - 32x + 9x - 12 = 0$
Rearranging the given equation, we get;
$\Rightarrow 8x\left( {3x - 4} \right) + 3\left( {3x - 4} \right) = 0$
So, we get $\left( {3x - 4} \right)\left( {8x + 3} \right) = 0$
$\Rightarrow \left( {3x - 4} \right) = 0$
Therefore , $x = \dfrac{4}{3}$
And also, $\left( {8x + 3} \right) = 0$
Therefore, $x = - \dfrac{3}{8}$
By this also we get the same two values of $x$ and they are $\dfrac{4}{3}$ and $- \dfrac{3}{8}$.
According to the given options , the correct answer for this question is option $\left( D \right)$ i.e. none of these.

Note:
For solving such problems, always remember the standard trigonometric identities. Here in this question we have used, ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$. Along with the knowledge of standard trigonometric formulae, some basic knowledge of algebraic rules and operations also eases the calculation.