In a ABC - 2 casin A - B - C -2 is equal toA- a2-b2-c2B- c2-a2-b2C- b2-c2-a2D- c2-a2-b2

The correct option is B We have A+C=π B And A+B+C/2=π/2 B ⇒ sin(A+B+C/2)=cos B ∴ 2accos B=2ca(c^2+a^2 b^2/2ac)=c^2+a^2 b^2 ...

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Global Maxima

In a ABC , 2 ...

Question

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

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

2 a c \sin (\frac{A - B + C}{2})

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

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

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

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

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)

△ A B C, \frac{s i n (A - B)}{s i n (A + B)} =

|\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}

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