If cosA=x/(y+z),cosB=y/(z+x),cosC=z/(x+y),show that tan^2A/2=tan^2B/2=tan^2C/2=1?

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Answer 1 · 2017-10-24T17:14:09

Taking $A=alpha,B=beta,C=gamma$

We have

$cosalpha=x/(y+z)$

$=>1/cosalpha=(y+z)/x$

By dividendo and componendo we get

$(1-cosalpha)/(1+cosalpha)=(y+z-x)/(x+y+z)$

$=>(2sin^2(alpha/2))/(2cos^2(alpha/2))=(y+z-x)/(x+y+z)$

$=>tan^2(alpha/2)=(y+z-x)/(x+y+z)$

Similarly

$tan^2(beta/2)=(x+z-y)/(x+y+z)$

And

$tan^2(gamma/2)=(x+y-z)/(x+y+z)$

Hence

$tan^2(alpha/2)+tan^2(beta/2)+tan^2(gamma/2)=(x+y+z)/(x+y+z)=1$