Question

Number of value(s) of $x$ which satisfy the equation ${\tan ^{ - 1}}\left( {2x - 1} \right) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {2x + 1} \right) = {\tan ^{ - 1}}4x$ is

Answer 1

Hint : Apply inverse trigonometric rules and also use basic concepts of trigonometry to simplify the equation so that you can compute all the values that satisfy the equation.

Complete step-by-step answer :
In this question the equation is given as,
${\tan ^{ - 1}}\left( {2x - 1} \right) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {2x + 1} \right) = {\tan ^{ - 1}}4x$
We have to find the number of values which satisfy the given equation.
Therefore, ${\tan ^{ - 1}}\left( {2x - 1} \right) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {2x + 1} \right) = {\tan ^{ - 1}}4x$
On transforming the value ${\tan ^{ - 1}}x$ from left hand side to right hand side we get,
${\tan ^{ - 1}}\left( {2x - 1} \right) + {\tan ^{ - 1}}\left( {2x + 1} \right) = {\tan ^{ - 1}}4x - {\tan ^{ - 1}}x$
Now applying the formula of ${\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x \pm y}}{{1 \pm xy}}} \right)$ on the both side in the above equation we get,
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2x - 1 + 2x + 1}}{{1 \pm (2x - 1)(2x + 1)}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{4x - x}}{{1 + 4x \times x}}} \right)$
On simplify the above equation we get,
${\tan ^{ - 1}}\left( {\dfrac{{4x}}{{4{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{3x}}{{-1 + 4{x^2}}}} \right)$
On calculating the equation by applying the inverse trigonometric rules we get,
$\dfrac{1}{x} = \dfrac{{3x}}{{-1 + 4{x^2}}}$
On cross multiplying the above equation we get,
$-1 + 4{x^2} = 3{x^2}$
On simplifying the above equation we get, $x = \pm 1$
Hence there are two values of $'x'$ which satisfies the equation is $\pm 1$ .

Note : In this type of question, you should make use of inverse trigonometric formula such as, ${\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y = \dfrac{{x \pm y}}{{1 \pm y}}$ , Also use $\tan \left[ {{{\tan }^{ - 1}}\theta } \right] = \theta$