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प्रश्न
In ΔABC, A + B + C = π show that
cos A + cos B – cos C = `4cos "A"/2 cos "B"/2 sin "C"/2 - 1`
उत्तर
L.H.S. = cos A + cos B – cos C
= `2cos(("A" + "B")/2) cos(("A" - "B")/2) - [1 - 2sin^2("C"/2)]`
= `2cos(pi/2 - "C"/2)cos(("A" - "B")/2) - 1 + 2sin^2("C"/2)` ...[∵ A + B + C = π]
= `2sin("C"/2)cos(("A" - "B")/2) + 2sin^2("C"/2) - 1`
= `2sin("C"/2)[cos(("A" - "B")/2) + sin^2("C"/2)] - 1`
= `2sin("C"/2)[cos(("A" - "B")/2) + sin{pi/2 - (("A" + "B")/2)}] - 1` ...[∵ A + B + C = π]
= `2sin("C"/2)[cos(("A" - "B")/2) + cos(("A" + "B")/2)] - 1`
= `2sin("C"/2) xx 2cos[(("A" + "B")/2 + ("A" - "B")/2)/2]*cos[(("A" + "B")/2 - ("A" - "B")/2)/2] - 1`
= `2sin("C"/2) xx 2cos("A"/2)*cos("B"/2) - 1`
= `4cos("A"/2)*cos("B"/2)*sin("C"/2) - 1`
= R.H.S.
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