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Question

tan1x+tan1y=tan1x+y1xy, xy<1
=π+tan1x+y1xy, xy>1.


tan1 {x(x+y+z)yz}+tan1 {y(x+y+z)zx}+tan1 {z(x+y+z)xy}=

A
π
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B
4π
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C
3π
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D
2π
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Solution

The correct option is A π
(a) Put x+y+z=r
tan1(rxyz)+tan1(ryzx)
=tan1r/xyz.(x+y)(1r/z)
=tan1(rz)/(xy).(x+y)(x+y)
=π+tan1{(rzxy)}
=πtan1(rzxy)
tan1(rzyz)+tan1(ryzx)+tan1(rzxy)=π


(b) We have to prove that
tan1yzxr+tan1zryr+tan1xyzr=π2
or tan1(z/r)[(x2+y2)/(xy)]1(z2/r2)=π2tan1xyzr
or tan1(zr/xy)(x2+y2)x2+y2=cot1xyzr
or ran1zrxy=tan1zrxy, which is true.

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