if A,B and C are the interior angle of a triangle ABC, show that
(1) sin (B+C/2)=cos (A/2)
(2) cosec (A+B/2)=sec (C/2)
(3)sec (C+A/2)=cosec (B/2)
Answers
Answered by
2
we know that
sum of the interior angle of a triangle is 180
A +B+C=180
C+B =180-A
C+B/2=,90-*A/2
TAKING SIN ON BITH SIDE
SIN(C+B/2)= SIN(90-A/2
SIN (C+B/2)=cos A/2
similarly next two u can prove
sum of the interior angle of a triangle is 180
A +B+C=180
C+B =180-A
C+B/2=,90-*A/2
TAKING SIN ON BITH SIDE
SIN(C+B/2)= SIN(90-A/2
SIN (C+B/2)=cos A/2
similarly next two u can prove
Similar questions