Prove that- 1-tan 1 2 Atan 1 2 B - 2c a-b-c-

1 tan 1 2 #160; Atan 1 2 #160; B =1 √(( ( s b ) ( s c ) #160; s( s a ) #160; #160; ) #160; ) .√(( ( s c ) ( s a ) #160; s( s b ...

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Circular Measurement of Angle

Prove that: ...

Question

Prove that: $1 - tan \frac{1}{2} A tan \frac{1}{2} B = \frac{2 c}{(a + b + c)}$ .

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\frac{c}{a + b} = \frac{1 - tan \frac{1}{2} A tan \frac{1}{2} B}{1 + tan \frac{1}{2} A tan \frac{1}{2} B}

\frac{tan \frac{1}{2} A}{(a - b) (a - c)} + \frac{tan \frac{1}{2} B}{(b - c) (b - a)} + \frac{tan \frac{1}{2} C}{(c - a) (c - b)} = \frac{1}{S}

In any $Δ A B C,$ prove that

$\frac{(b - c)}{b + c} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$

t a n \frac{1}{2} A = \frac{5}{6}

t a n \frac{1}{2} B = \frac{20}{37}

t a n \frac{1}{2} C,

a + c = 2 b .

In $Δ A B C,$ prove that

$\frac{(b - c)}{b + c} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$

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