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Trignometry

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2 Apr 2008 17:30:44 IST

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Please explain it once more

None

If A,B and C,are the angles of a triangle,show that

tan²A/2+tan²B/2+tan²C/2

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ram kumar

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Joined: 14 Aug 2007

Posts: 407

2 Apr 2008 17:55:03 IST

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here is my proof:::::::::

let x,y,z>0

$x^2+y^2\;\geq\;2xy\;\;\;(A.M \geq G.M)$

$y^2+z^2\;\geq\;2yz$

$z^2+x^2\;\geq\;2xz$

adding the above inequalities we get......

$x^2+y^2+z^2\;\geq\;xy+yz+xz$ ..........(1)

since $tan\frac{A}{2},tan\frac{B}{2},tan\frac{C}{2}>0$

we can substitute $x=tan\frac{A}{2},y=tan\frac{B}{2},z=tan\frac{C}{2}$

thus we get from (1)......

$tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}\geq tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}$

thus

$tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}\geq 1$

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