Question

10. From the above figure, cosec C =
2po
A
3
C
B
BC/AC
AB/AC
O AC/AB​

Answer 1

if triangle is right angled at b then b= 90°

so acc to question

(sec.a)(cosec.c) - (tan.a)(cot.c)/(sin.b)

if angle a is taken into consideration then

sec.a = ac/ab

tan.a = bc/ab

and

sin.b = sin(90°) = 1

so equation becomes

(ac/ab)(cosec.c) - (bc/ab)(cot.c)

now if angle c is taken into consideration

cosec.c = ac/bc

cot.c = ab/bc

so filling values

(ac × ac)/(ab × bc) - (bc × ab)/(ab × bc)

{ (ac^2) - (ab×bc) }/ (ab)(bcl

Answer 2

AC/AB

cosec=hypotenuse/adjacent