Question

in triangle abc prove that
sec(B+C/2)=CosecA/2​

Answer 1

Answer:

Hello friend hope this helps you and mark me as the brainiest

Step-by-step explanation:

<a + <b + <c = 180 ......[ angle sum property]                                                  Dividing both sides by 2                           <a + <b + <c / 2 = 180/2                                <b + <c /2 = 90 - <c/2                                 Sec<b + <c /2 = sec 90- <a/2.                  Sec<b + <c/2 = cosec<a/2.                       Hence proved.......

Answer 2

Answer:

I hope it will help you.

Step-by-step explanation:

In?ABC A+B+C =180

B+C=180-A

NOW SEC(B+C2)= SEC( 180-A /2)= SEC (90-A/2)= COSEC(A2)

THAT'S ALL BRO