Question

Triangle abc, prove:
cos(a+c) + cosb=0

Answer 1

Answer:

cos(a+c) + cosb=0

taking lhs

lhs=cos(a+c) + cosb

we kno that triangle is completely 180°

and in the trigonometry has right angle triangle

if we suppose b is right angle

b=90°

now,

we have

a+c=90°

here

we suppose

a=30°

c=60°

now,

=cos(a+c) + cosb

=cos(30+60) + cos90

=cos(90) + cos90

=0+0. ------------------(cos90=0)

=0

hence proved