Question

if a,b,c are interior angles of a triangle abc then show that cosec (angleb+anglec÷2)=secanglea÷2

Answer 1

If A,B,C are the interior angles of a triangle,
Then A+B+C=180°.
B+C=180-A
B+C÷2=180-A÷2
=90-A/2

Cosec(B+C÷2)=Cosec (90-A/2)
=Sec(A/2)