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 $