solve tan inverse 2x + tan inverse 3x = pie/4. (in the end we will have x = -1 and x=1/6 then in ncert x = -1 is rejected but if we put x=-1 in L.H.S after adding them it satisfies the equation ??)
tan-1(2x) + tan-1(3x) = (π/4) (1)
So using tan-1A + tan-1B = tan-1{(A+B)/(1- AB)}
So tan-1(2x) + tan-1(3x) = tan-1 {( 2x+3x)/(1-6x2 )}
So taking tan on both sides we get ( 2x+3x)/(1-6x2 ) = 1
6x2 + 5x -1 = 0
So x = -1 and x = 1/6
But if you put x = -1 in (1), we get LHS as negative, so only 1/6 is the answer.
Hope this would have cleared your doubt.
So using tan-1A + tan-1B = tan-1{(A+B)/(1- AB)}
So tan-1(2x) + tan-1(3x) = tan-1 {( 2x+3x)/(1-6x2 )}
So taking tan on both sides we get ( 2x+3x)/(1-6x2 ) = 1
6x2 + 5x -1 = 0
So x = -1 and x = 1/6
But if you put x = -1 in (1), we get LHS as negative, so only 1/6 is the answer.
Hope this would have cleared your doubt.