STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{2x + 3x}}{{1 - (2x)\,(3x)}}} \right) = \frac{\pi }{4}\,$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{5x}}{{1 - 6{x^2}}}} \right)\, = {\tan ^{ - 1}}(1)$

$ \Rightarrow \,\,\frac{{5x}}{{1 - 6{x^2}}} = 1\,$

$ \Rightarrow \,\,1 - 6{x^2} = 5x$

$\, \Rightarrow \,\,6{x^2} + 5x - 1 = 0$

$ \Rightarrow \,\,(x + 1)\,\left( {x - \frac{1}{6}} \right) = 0\,$

$ \Rightarrow \,\,x = - 1,\,\,\frac{1}{6}$

But $-1$ does not hold.

Trick : Check with the options. 

Obviously the equation holds for $x = \frac{1}{6}$, but not for $-1$.