In a triangle ABC- 2ac sin12 A - B - C -

The correct option is D c^2 + a^2 b^2 We know that, A+B+C=π ⇒ A+C=π B ⇒ A+C B=A+B+C 2B=2( π #160; 2 B ) So, here #160; 2acsin ...

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Trigonometric Ratios of Half Angles

In a triangle...

Question

In a triangle $A B C, 2 a c sin \frac{1}{2} (A - B + C) =$

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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|\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}

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

In a $△ A B C, 2 c a s i n (\frac{A - B + C}{2})$ is equal to

△ 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}

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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