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In any DeltaABC, prove that ((a-b))/c"co...

In any `DeltaABC`, prove that `((a-b))/c"cos"C/2="sin"((A-B))/2`

A

`"sin"((A-B))/(2)`

B

`tan((A-B)/(2))`

C

`"sin"((A+B))/(2)`

D

`"tan"((A+B))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
By the sine rule ,we have
`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)=k`
`rArra=ksinA,b=ksinBandc=ksinC`. . . . (i)
`((a-b))/(b)"cos"(C)/(2)=((ksinA-ksinB))/(ksinC)"cos"(C)/(2)`
`=(k(sinA-sinB))/(ksinC)*"cos"(C)/(2)` [fromEq . (i)]
`=((sinA-sinB))/(sinC)*(C)/(2)`
`=(2"cos"((A+B))/(2)"sin"*((A-B))/(2))/(2"sin"(C)/(2)"cos"(C)/(2))"cos"(C)/(2)`
`[because(sinC-sinD)=2"cos"((C+D))/(2)"sin"((C-D))/(2)]`
`=(cos((pi)/(2)-(C)/(2))*"sin"((A-B))/(2))/("sin"(C)/(2))`
`[because(A+B+C)/(2)=(pi)/(2)rArr(A=B)/(2)=((pi)/(2)-(C)/(2))]`
`=("sin"(C)/(2)*"sin"((A-B))/(2))/("sin"(C)/(2))="sin"((A-B))/(2)`
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