results involving the angles of a triangle (1 Viewer)

[AFGHAN22]

if a b and c are the angles of a triangle prove that

tan2a+tan2b+tan2c=tan2atan2btan2c

cot(a/2) + cot(b/2) + cot (c/2)=cot(a/2)cot(b/2)cot(c/2)

please help guys, these have been bothering me for the last two days

 

[Roobs]

http://www.classontheweb.com/samples/getsetgo/cbse/clas11/maths/chap9h.htm

for the first one

 

[Riviet]

Starting with a + b + c = pi, (angle sum of triangle)
-a/2 - b/2 - c/2 = -pi/2
(pi/2 - a/2) + (pi/2 - b/2) - c/2 = pi/2
tan[(pi/2 - a/2) + (pi/2 - b/2)] = tan(c/2 + pi/2)
= tan(pi/2 -(-c/2))
= cot(-c/2)
=-cot(c/2)
LHS=
cot(a/2) + cot(b/2)
1 - cot(a/2).cot(b/2)

cot(a/2) + cot(b/2)=-cot(c/2){1 - cot(a/2).cot(b/2)}
cot(a/2) + cot(b/2)=-cot(c/2) + cot(a/2).cot(b/2).cot(c/2)
cot(a/2) + cot(b/2) + cot(c/2) = cot(a/2).cot(b/2).cot(c/2) #