If a-b-c-0 then 1-b2-c2-a2-1-c2-a2-b2-1-a2-b2-c2 is equal to

The correct option is D 0Given a+b+c=0 ⇒ b+c= a Squaring on both sides ; ⇒ b^2+c^2+2bc=a^2 ⇒ b^2+c^2 a^2= 2bc Similarly c^2+a^2 b^2= 2a ...

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Quadratic Formula

If a+b+c=0 ...

Question

If $a + b + c = 0$ then $\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}$ is equal to

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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 R^{+}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

a = 1

b = - 1

c = 0

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

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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