`(sin2A+sin2B+sin2C)/(sinA+sinB+sinC)i se q u a lto` `8sinA/2sinB/2sinC/2` (b) `8cosA/2cosB/2cosC/2` `8tanA/2tanB/2tanC/2` (d) `8cot(A/2B)/2cotC/2`

Options

A. `8 sin((A)/(2))sin((B)/(2))sin((C)/(2))`
B. `8cos((A)/(2))cos((B)/(2))cos((C)/(2))`
C. `8tan((A)/(2))tan((B)/(2))tan((C)/(2))`
D. `8cot((A)/(2))cot((B)/(2))cot((C)/(2))`

Answer

A

Explanation

`(sin2A+sin2B+sin2C)/(sinA+sinB+sinC)=(4sinAsinBsinC)/(4cos((A)/(2))cos((B)/(2))cos((C)/(2)))`
`=8sin((A)/(2))sin((B)/(2))sin((C)/(2))[because sinA=2sin((A)/(2))cos((A)/(2))]`