Question

If a b c are interior angles of a triangle abcthen prove that seca =a+c/2=cosecb/2

Answer 1

a+b+c = 180degree. [ by angle sum property ]
by dividing each term by 2 , we get
a/2 +b/2 +c/2 = 180 degree
a/2 + c/2 = 90 degree -b/2
(a+c)/2 = 90 degree - a/2
now, multiplying both sides by sec , we get,
sec(a+c)/2 = sec(90 degree - a/2)
since sec(90 degree - x) = cosec x,
therefore,
sec(a+c)/2 = cosec a/2
[hence, proved]