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Question
In ΔABC prove that `sin "A"/(2). sin "B"/(2). sin "C"/(2) = ["A(ΔABC)"]^2/"abcs"`
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Solution
L.H.S.
= `sin "A"/(2). sin "B"/(2). sin "C"/(2)`
`= sqrt(((s - b)(s - c))/"bc") .sqrt(((s - a)(s - c))/"ac") . sqrt(((s -a )(s - b))/"ab"`
`= sqrt(((s - a)^2(s - b)^2(s - c)^2)/(a^2b^2c^2)`
`= ((s - a)(s - b)(s - c))/"abc"`
`= (s(s - a)(s - b)(s - c))/"abcs"`
`= (["A(ΔABC")]^2/"abcs" ...[∵ "A(ΔABC") = sqrt(s(s - a)(s - b)(s - c))]`
= R.H.S.
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