If a+b+c=2π,then prove that SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2

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Aditya Gupta · Answer

the question is wrong and the two sides are noe not identically equal as lhs is greater than equal to 1 while rhs is less than equal to 1. however, for particular value like a=b=0 and c=2pi, this might indeed hold true

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If a+b+c=2π,then prove that SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2

If a+b+c=2π,then prove that

SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2

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If a+b+c=2π,then prove that SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2

If a+b+c=2π,then prove that SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2

1 Answers

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If a+b+c=2π,then prove that

SecA/2×secB/2×secc/2=cosA/2×cosB/2×cosc/2