In - A B C with usual notations 2a c sin 1 2 A - B - C is equal to-

The correct option is A c ^ 2 + a ^ 2 b ^ 2 2acsin 1 2 ( A B+C ) =2acsin 1 2 ( A+B+C 2B ) =2acsin 1 2 ( π 2B ) =2acsin( π / ...

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Logarithmic Differentiation

In Δ A B C ...

Question

In $Δ A B C$ with usual notations $2 a c sin \frac{1}{2} (A - B + C)$ is equal to-

a^{2} + b^{2} - c^{2}

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c^{2} - a^{2} - b^{2}

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b^{2} - c^{2} - a^{2}

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c^{2} + a^{2} - b^{2}

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△ A B C

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

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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