Question

if A,B,C, are angles of triangle ABC then show that sec[A+C]/2=cosec [B/2]​

Answer 1

Step-by-step explanation:

the answer is B and you will of course pass

Answer 2

Step-by-step explanation:

Sum of all angles of a triangle=>

A+B+C=180°.

A+C=180-B

Divi$e both sides by 2:

A+C/2=180°-B/2

A+C/2=180°/2-B/2

A+C/2=90°-B/2

Add sec to both side:

sec(A+C/2)=sec(90°-90°-B/2)

sec(A+B/2)=cosec(B/2).