Trigonometry..

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In a \Delta ABC, let

x=\tan\left(\frac{B-C}{2}\right)\tan\left(\frac{A}{2}\right)

y =\tan\left(\frac{C-A}{2}\right)\tan\left(\frac{B}{2}\right)

z =\tan\left(\frac{A-B}{2}\right)\tan\left(\frac{C}{2}\right)

Prove that x+y+z+xyz = 0

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aieeee ·2009-07-26 21:04:30

ACTUALLY,its a formula,which goes this way: tan(B-C/2) = b-c/b+c hope,u cn solve it now.

( derirvation of the formula:

b-c/b+c = sinB-sinC/sinB+sinC ,
= (2 sin B-C/2 *cos B+C/2) / ( 2 sin B+C/2*cos B-C/2)
= (tan B-C/2) / (tan B+C/2)
now,putting B+C=âˆ© - A, U GET THE FORMULA

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Devil ·2009-07-27 06:51:45

Napier's Analogy....

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Basic Integral Calculus Operators Symbols Matrix Cases

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Trigonometry..

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