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Question

In a triangle ABC we define
x=tanBC2tanA2,y=tanCA2tanB2 and z=tanAB2tanC2
Then the value of x+y+z (in terms of x,y,z) is


A

xyz

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B

-xyz

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C

2xyz

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D

none

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Solution

The correct option is B

-xyz


tanBC2=bcb+ccotA2, Napier's analogyx1=tanBC2tanA2=bcb+cxb+xc=bcc(x+1)b(1x)1+x1x=bcSimilarly 1+y1y=ca and 1+z1z=ab(1+x)(1+y)(1+z)(1x)(1y)(1z)=bc.ca.ab=1(1+x)(1+y+z+yz)(1x)(1yz+yz)=11+y+z+yz+x+xy+xz+xyz=1yz+yzx+xy+xzxyz2(x+y+z)=2xyz(x+y+z)=xyz


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