. - tan(nπ+θ)=tanθ . - tan(−θ)=−tanθ Calculations: It is given that
tanx
2
=
tany
3
=
tanz
5
=k (say). ∴tanx=2k,tany=3k and tanz=5k . It is also given that x+y+z=π . ⇒x+y=π−z ⇒tan(x+y)=tan[π+(−z)] ⇒
tanx+tany
1−tanxtanx
=tan(−z)=−tanz ⇒tanx+tany=−tanz+tanxtanytanz ⇒tanx+tany+tanz=tanxtanytanz Substituting the values in terms of k from the above result, we get: 2k+3k+5k=(2k)(3k)(5k) ⇒10k=30k3 ⇒k2=