Prove that- b-c a cos 2 1 2 A- c-a b cos 2 1 2 B- a-b c cos 2 1 2 C -0-

∴ L.H.S+ s abc [ ( s a ) ( b c ) #160; ] ∴ L.H.S= s abc [ ( s a ) ( b c ) +( s b ) ( c a ) +( s c ) ( a b ) #160; ] = s abc [ { b c+c a+a ...

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Chain Rule of Differentiation

Prove that: ...

Question

Prove that: $\frac{b - c}{a} {cos}^{2} \frac{1}{2} A + \frac{c - a}{b} {cos}^{2} \frac{1}{2} B + \frac{a - b}{c} {cos}^{2} \frac{1}{2} C = 0.$

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a {cos}^{2} \frac{C}{2} + c {cos}^{2} \frac{A}{2} = \frac{3 b}{2}

△ A B C

b {cos}^{2} \frac{A}{2} + a {cos}^{2} \frac{B}{2} = \frac{3 c}{2}

\frac{a + c}{2 c - a} + \frac{b + c}{2 c - 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}

{(\frac{2^{a}}{2^{b}})}^{a + b} \times {(\frac{2^{b}}{2^{c}})}^{b + c} \times {(\frac{2^{c}}{2^{a}})}^{c + a} = 1

a, b, c

△ A B C

A . P .

a {cos}^{2} \frac{C}{2} + c {cos}^{2} \frac{A}{2} = \frac{3 b}{2}

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