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?

Question

2537 views around the world

You can reuse this answer
Creative Commons License

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 #