In triangle `A B C ,(sinA+sinB+sinC)/(sinA+sinB-sinC)` is equal to `tanA/2cotB/2` (b) `cotA/2tanB/2` `cotA/2cotB/2` (d) `tanA/2tanB/2`

Options

A. `tan((A)/(2))cot((B)/(2))`
B. `cot((A)/(2))tan((B)/(2))`
C. `cot((A)/(2))cot((B)/(2))`
D. `tan((A)/(2))tan((B)/(2))`

Answer

C

Explanation

Denominator `=sinA+sinB-sinC`
`=2sin ((A+B)/(2))cos((A-B)/(2))-2sin((C)/(2))cos((C)/(2))`
`=cos((C)/(2))[cos((A-B)/(2))-sin((C)/(2))]`
`=2cos((C)/(2))[cos((A-B))/(2))-cos((A+B)/(2))]`
`=2cos((C)/(2))[2sin((A)/(2))sin((B)/(2))]`
`=4sin((A)/(2))sin((B)/(2))cos((C)/(2))` .
Also `sinA+sinB+sinC=4cos((A)/(2))cos((B)/(2))cos((C)/(2))`
`rArr (sinA+sinB+sinC)/(sinA+sinB-sinC)=cot((A)/(2))cot((B)/(2))`.