The correct options are
A x=0,n=0
B x=1,n=1
C x=−1,n=−1
tan−1(x)+tan−1(2x)
=tan−1(3x1−2x2)
=nπ−tan−13x
Hence, 3x1−2x2=tan(nπ−tan−1(3x))
Now for all nϵI+ we get
3x1−2x2=tan(nπ−tan−1(3x))
=−tan(tan−1(3x))
=−3x
Hence
3x1−2x2=−3x
3x[11−2x2+1]=0
3x[2−2x21−2x2]=0
6x[1−x21−2x2]=0
Hence x=0 and x2−1=0
x=±1.
Hence for all nϵ{0,1,2,...N} we get x=−1,0,1.
Now for nϵI− we get
3x1−2x2=−tan−1(nπ+tan−1(3x))
3x[2−2x21−2x2]=0
6x[1−x21−2x2]=0
Hence for nϵ{−1,−2,−3...−N} x={−1,0,1}.