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Question

Solve tan1x+tan12x=nπtan13x,nϵI,nϵR,for n and x

A
x=0,n=0
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B
x=1,n=1
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C
x=1,n=1
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D
x=2,n=2
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Solution

The correct options are
A x=0,n=0
B x=1,n=1
C x=1,n=1
tan1(x)+tan1(2x)
=tan1(3x12x2)
=nπtan13x
Hence, 3x12x2=tan(nπtan1(3x))
Now for all nϵI+ we get
3x12x2=tan(nπtan1(3x))
=tan(tan1(3x))
=3x
Hence
3x12x2=3x
3x[112x2+1]=0
3x[22x212x2]=0
6x[1x212x2]=0
Hence x=0 and x21=0
x=±1.
Hence for all nϵ{0,1,2,...N} we get x=1,0,1.
Now for nϵI we get
3x12x2=tan1(nπ+tan1(3x))
3x[22x212x2]=0
6x[1x212x2]=0
Hence for nϵ{1,2,3...N} x={1,0,1}.

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