Question

x-y=Tan (2x/1-x^2)+tan^-1(3x-x^2/1-3x^2) tan^-1 4x-4x^3/1-6x^2-x^2​

Answer 1

Answer:

The question is incomplete.

Step-by-step explanation:

$x - y\: = \tan^{ - 1}( \frac{2x}{1 - {x}^{2} } ) + \tan^{ - 1}( \frac{3x - {x}^{2} }{1 - 3 {x}^{2} } ) + \tan^{ - 1}( \frac{4x - 4 {x}^{3} }{1 - 6 {x}^{2} - {x}^{2} } ) \\ \tan^{ - 1} (2x) + \tan^{ - 1} (3x) + \tan^{ - 1} (4x) \\ \tan^{ - 1} ( \frac{2x + 3x + 4x - 2x.3x.4x}{1 - 2x.3x - 3x.4x - 4x.2x} ) =\tan^{ - 1}( \frac{9x - 24x ^{3} }{1 - 26x^{2} } )$