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In a DeltaABC, cosecA(sinBcosC+cosBsinC)...

In a `DeltaABC`, cosecA(sinBcosC+cosBsinC)` is equal to

A

`(c)/(a)`

B

`(a)/(c)`

C

1

D

`(c)/(ab)`

Text Solution

Verified by Experts

The correct Answer is:
C
We have ,`cosecA(sinBcosC+cosBsinC)`
`=((sinB)/(sinA)"cos"C+(sinC)/(sinA)"cos"B)`
`=((b)/(a)"cos"C+(c)/(a)"cos"B)=(a)/(a)=1`
`[because a=bcosC+ccisB]`
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