If a - b - c - 2s- then s - a2 - s - b2 - s - c2 - s2a2 - b2 - c2 is equal to

The correct option is D 1 Given, a + b + c = 2s = (s a)^2 + (s b)^2 + (s c)^2 + s^2a^2 + b^2 + c^2 = s^2 + a^2 2sa + s^2 + b^2 2s ...

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Composition of Function

If a + b + ...

Question

If $a + b + c = 2 s$ , then $\frac{(s - a)^{2} + (s - b)^{2} + (s - c)^{2} + s^{2}}{a^{2} + b^{2} + c^{2}}$ is equal to

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

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Similar questions

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 = 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}}

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

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

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

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