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प्रश्न
If A + B + C = 2s, then prove that sin(s – A) sin(s – B)+ sin s sin(s – C) = sin A sin B
उत्तर
Now sin(s – A) sin(s – B) = `1/2 {cos[("s" - "A") - ("s" - "B")] - cos[("s" - "A") + ("s" - "B")]}`
= `1/2cos("s" - "A" - "s" + "B") - cos[2"s" - ("A" + "B")]`
= `1/2 {cos("A" - "B) - cos"C"}` .....[∴ cos(A – B) = cos(B – A)]
Again sin s sin s – C = `1/2[cos"C" - cos("A" + "B")`
So, L.H.S = `1/2 {cos("A" - "B") - cos"C" + cos"C" - cos("A" + "B")}`
= `1/2 [cos("A" - "B") - cos("A" + "B")`
= `1/2 [2sin"A" sin"B"]`
= sin A sin B
= R.H.S
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